Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Resistencia de los materiales (ejercicios- equilibrio estático, diagrama de c...josekat_17
This document provides calculations and diagrams for determining reactions and moments in statically determinate beams with distributed loads. It includes 6 example problems showing the process of: 1) establishing static equilibrium equations to solve for support reactions, 2) calculating shear force and bending moment diagrams, and 3) determining maximum bending moments. Diagrams are drawn and key values calculated at each step.
1) O documento apresenta fórmulas para cálculo de momento de inércia, momento estático e módulo de resistência para diferentes seções transversais.
2) São apresentados exemplos de cálculo de tensões longitudinais em vigas sob flexão, incluindo determinação de diagramas de esforços internos, posição da linha neutra e momentos fletores.
3) Pede-se para calcular a máxima tensão longitudinal em uma barra cilíndrica sob ação de momento fletor uniforme.
This document provides an overview of OpenStack APIs and the WSGI (Web Server Gateway Interface) that powers them. It begins with an introduction to WSGI and how OpenStack services are implemented as WSGI applications. It then demonstrates how the OpenStack APIs can be accessed via libraries like novaclient or directly with HTTP requests. Code examples are provided showing how to authenticate against Keystone and retrieve images using urllib2. The document concludes with explanations of how WSGI, WebOb, and Paste are used to implement the OpenStack "web stack".
Este documento descreve os conceitos de flexão simples em vigas. A flexão simples ocorre quando uma viga é submetida apenas a um momento fletor variável ao longo dela, gerando tensões normais e tangenciais. As tensões normais seguem a equação de Euler e têm distribuição linear. Já as tensões tangenciais surgem devido à variação do momento fletor e seguem a equação de Jourawsky, com distribuição não linear na seção transversal. Exemplos demonstram o cálculo destas tensões em diferentes formatos de
Slide show developed as part of school project. My contribution to group, in addition to engineering and circuit design, was creating and organizing a Power Point presentation to summarize our design.
This document discusses the history of chocolate production. It details how cocoa beans are harvested from cocoa trees and then fermented, dried, roasted, and ground into chocolate liquor. The liquor is then further processed through conching and tempering to produce chocolate in its familiar solid form.
Resistencia de los materiales (ejercicios- equilibrio estático, diagrama de c...josekat_17
This document provides calculations and diagrams for determining reactions and moments in statically determinate beams with distributed loads. It includes 6 example problems showing the process of: 1) establishing static equilibrium equations to solve for support reactions, 2) calculating shear force and bending moment diagrams, and 3) determining maximum bending moments. Diagrams are drawn and key values calculated at each step.
1) O documento apresenta fórmulas para cálculo de momento de inércia, momento estático e módulo de resistência para diferentes seções transversais.
2) São apresentados exemplos de cálculo de tensões longitudinais em vigas sob flexão, incluindo determinação de diagramas de esforços internos, posição da linha neutra e momentos fletores.
3) Pede-se para calcular a máxima tensão longitudinal em uma barra cilíndrica sob ação de momento fletor uniforme.
This document provides an overview of OpenStack APIs and the WSGI (Web Server Gateway Interface) that powers them. It begins with an introduction to WSGI and how OpenStack services are implemented as WSGI applications. It then demonstrates how the OpenStack APIs can be accessed via libraries like novaclient or directly with HTTP requests. Code examples are provided showing how to authenticate against Keystone and retrieve images using urllib2. The document concludes with explanations of how WSGI, WebOb, and Paste are used to implement the OpenStack "web stack".
Este documento descreve os conceitos de flexão simples em vigas. A flexão simples ocorre quando uma viga é submetida apenas a um momento fletor variável ao longo dela, gerando tensões normais e tangenciais. As tensões normais seguem a equação de Euler e têm distribuição linear. Já as tensões tangenciais surgem devido à variação do momento fletor e seguem a equação de Jourawsky, com distribuição não linear na seção transversal. Exemplos demonstram o cálculo destas tensões em diferentes formatos de
Slide show developed as part of school project. My contribution to group, in addition to engineering and circuit design, was creating and organizing a Power Point presentation to summarize our design.
This document discusses the history of chocolate production. It details how cocoa beans are harvested from cocoa trees and then fermented, dried, roasted, and ground into chocolate liquor. The liquor is then further processed through conching and tempering to produce chocolate in its familiar solid form.
El documento proporciona instrucciones en 7 pasos para completar un proceso de carga en un programa de inventario. Los pasos incluyen acceder al menú principal a través del icono de inventario, seleccionar la opción de carga en el menú de inventario, completar una ventana emergente con descripción, fecha y detalles de los cargos, y finalmente registrar y aceptar la transacción.
The passive voice is used when the subject of the sentence receives or is affected by the action rather than performing the action. The passive is often used to focus on the recipient of the action rather than the performer, and the agent (performer) is sometimes omitted if it is unknown, unimportant, or obvious from context. The passive has an impersonal and formal style and is commonly used in notices and announcements, while the active voice is more informal and uses pronouns like "you", "we", and "they" to refer to people.
The document discusses the benefits of exercise for mental health. Regular physical activity can help reduce anxiety and depression and improve mood and cognitive functioning. Exercise causes chemical changes in the brain that may help boost feelings of calmness, happiness and focus.
This document provides solutions to problems from chapters 1-9 of a mechanical engineering design textbook. For each problem, it lists the problem number, chapter, and solutions including calculated values, equations and brief descriptions. The solutions are technical in nature and include terms like stress, strain, force and calculations of values for these parameters. Over 40 problems are summarized with calculated results.
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
In the 1930s, women wore wool suits with padded shoulders and knee-length flared skirts, along with hats for evening wear. Men wore suits for work with wide-legged pants that were tight at the ankles, open-neck shirts without ties, pointy shoes, and hats, often with suspenders. Today's fashion has no single dominant style but many styles together that are totally different from each other but all coexist. The current fashion is considered more beautiful, colorful, and original than the styles of the 1930s.
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
This chapter discusses forcible entry tools and techniques used by firefighters. It covers selecting the appropriate cutting, prying, pushing/pulling and striking tools for specific applications. Some key tools discussed include axes, bolt cutters, saws and hydraulic tools. The chapter emphasizes selecting the proper tools for the job, using tools as intended and following safety procedures to avoid injuries. It also addresses maintaining and carrying tools properly. The overall goal is for firefighters to safely force entry through various barriers like doors, locks, windows and walls according to their department's standard operating procedures.
Los procesos organizacionales consisten en roles, relaciones, autoridades y responsabilidades para lograr objetivos de manera eficiente y eficaz. Estos procesos pueden ser formales o informales y comprenden la comunicación, el liderazgo, la toma de decisiones y la dirección dentro de una organización. La comunicación, la autoridad, el poder, el liderazgo y la toma de decisiones son elementos clave de los procesos organizacionales.
The document reports weight loss results over 18 weeks. It shows a person lost 16.3 kg between Week 1 and Week 18. The results are repeated to emphasize the 16.3 kg weight loss over the period.
This document contains example problems for the selection and design of ball and roller bearings. Problem 11-1 provides an example calculation to select a deep-groove ball bearing based on its rated load capacity and required design life. Problem 11-2 performs similar calculations to select an angular-contact ball bearing. Problem 11-3 extends this to the selection of a straight roller bearing. The remaining problems provide additional examples of selecting bearings based on load conditions, reliability requirements, and combined load considerations.
The document provides examples of calculating bearing load ratings (C10) using various parameters including design life (LD), design load (FD), reliability (R), and Weibull parameters. It also gives examples of selecting bearings from catalog ratings to meet load and reliability requirements. Overall reliability is calculated by multiplying the reliabilities of individual bearings.
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
This chapter discusses applying fatigue concepts from Chapter 7 to shaft design. It presents examples of calculating shaft diameters to satisfy deflection, distortion, and strength constraints. The chapter concludes by noting each design will differ in details and no single solution is presented for the open-ended problem of selecting bearings and designing attachments for a given shaft. Students are provided experience applying analysis to iteratively size shafts and assess adequacy of individual designs.
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
El documento proporciona instrucciones en 7 pasos para completar un proceso de carga en un programa de inventario. Los pasos incluyen acceder al menú principal a través del icono de inventario, seleccionar la opción de carga en el menú de inventario, completar una ventana emergente con descripción, fecha y detalles de los cargos, y finalmente registrar y aceptar la transacción.
The passive voice is used when the subject of the sentence receives or is affected by the action rather than performing the action. The passive is often used to focus on the recipient of the action rather than the performer, and the agent (performer) is sometimes omitted if it is unknown, unimportant, or obvious from context. The passive has an impersonal and formal style and is commonly used in notices and announcements, while the active voice is more informal and uses pronouns like "you", "we", and "they" to refer to people.
The document discusses the benefits of exercise for mental health. Regular physical activity can help reduce anxiety and depression and improve mood and cognitive functioning. Exercise causes chemical changes in the brain that may help boost feelings of calmness, happiness and focus.
This document provides solutions to problems from chapters 1-9 of a mechanical engineering design textbook. For each problem, it lists the problem number, chapter, and solutions including calculated values, equations and brief descriptions. The solutions are technical in nature and include terms like stress, strain, force and calculations of values for these parameters. Over 40 problems are summarized with calculated results.
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
In the 1930s, women wore wool suits with padded shoulders and knee-length flared skirts, along with hats for evening wear. Men wore suits for work with wide-legged pants that were tight at the ankles, open-neck shirts without ties, pointy shoes, and hats, often with suspenders. Today's fashion has no single dominant style but many styles together that are totally different from each other but all coexist. The current fashion is considered more beautiful, colorful, and original than the styles of the 1930s.
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
This chapter discusses forcible entry tools and techniques used by firefighters. It covers selecting the appropriate cutting, prying, pushing/pulling and striking tools for specific applications. Some key tools discussed include axes, bolt cutters, saws and hydraulic tools. The chapter emphasizes selecting the proper tools for the job, using tools as intended and following safety procedures to avoid injuries. It also addresses maintaining and carrying tools properly. The overall goal is for firefighters to safely force entry through various barriers like doors, locks, windows and walls according to their department's standard operating procedures.
Los procesos organizacionales consisten en roles, relaciones, autoridades y responsabilidades para lograr objetivos de manera eficiente y eficaz. Estos procesos pueden ser formales o informales y comprenden la comunicación, el liderazgo, la toma de decisiones y la dirección dentro de una organización. La comunicación, la autoridad, el poder, el liderazgo y la toma de decisiones son elementos clave de los procesos organizacionales.
The document reports weight loss results over 18 weeks. It shows a person lost 16.3 kg between Week 1 and Week 18. The results are repeated to emphasize the 16.3 kg weight loss over the period.
This document contains example problems for the selection and design of ball and roller bearings. Problem 11-1 provides an example calculation to select a deep-groove ball bearing based on its rated load capacity and required design life. Problem 11-2 performs similar calculations to select an angular-contact ball bearing. Problem 11-3 extends this to the selection of a straight roller bearing. The remaining problems provide additional examples of selecting bearings based on load conditions, reliability requirements, and combined load considerations.
The document provides examples of calculating bearing load ratings (C10) using various parameters including design life (LD), design load (FD), reliability (R), and Weibull parameters. It also gives examples of selecting bearings from catalog ratings to meet load and reliability requirements. Overall reliability is calculated by multiplying the reliabilities of individual bearings.
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
This chapter discusses applying fatigue concepts from Chapter 7 to shaft design. It presents examples of calculating shaft diameters to satisfy deflection, distortion, and strength constraints. The chapter concludes by noting each design will differ in details and no single solution is presented for the open-ended problem of selecting bearings and designing attachments for a given shaft. Students are provided experience applying analysis to iteratively size shafts and assess adequacy of individual designs.
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
This document provides calculations for determining braking forces and torques in a drum brake system. It gives equations and applies them to example problems, providing values for forces, pressures, and torques. In one example, a maximum pressure of 734.5 kPa is found on the right shoe for clockwise rotation. The total braking torque is calculated to be 422 N-m. In another example, a braking capacity of 1750 N-m is determined.
1. The document describes the design of a gear box including specifications, calculations of gear ratios and sizes, material selection, and shaft design.
2. Key specifications include delivering 27.6 hp at input speed of 1992 rpm and output speed of 103 rpm. Gears and bearings are designed to last over 13,000 hours.
3. Calculations determine the number of teeth for each gear and gear ratios. Materials including various grades of steel are selected for gears and shafts based on bending and wear stresses.
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
This document provides calculations and design considerations for sizing a flat belt drive system. It begins by giving parameters for the system including angular velocity ratio, nominal horsepower, pulley center distance, stiffness, and material properties. Initial calculations are shown to check for developed friction. The document then considers doubling the system size as a design task and provides the scaled calculations. It next shows a sample design process for selecting a belt width to meet the specified horsepower requirements. Finally, it solves for several values related to the maximum torque and slip conditions for the belt drive.
This document contains solutions to problems from Chapter 5 of an engineering textbook. Problem 5-3 calculates the torque and allowable twist in a torsion bar made of two springs in parallel. Problem 5-12 calculates the maximum deflection and stress in a beam loaded by two point loads. Problem 5-19 involves selecting the appropriate cross-sectional dimensions to achieve a required stiffness for a beam of given length.
Here are the steps to design and draw a flywheel for a four stroke four cylinder 133 kW engine running at 375 rpm with a diameter not exceeding 1 m:
Given:
Power of engine, P = 133 kW = 133000 W
Number of cylinders, n = 4
Speed of engine, N = 375 rpm
Maximum diameter, Dmax = 1 m
Step 1) Calculate the mean effective pressure (p):
p = (2*P)/(n*π*D^2*N)
p = (2*133000)/(4*π*(0.5)^2*375) = 7 bar
Step 2) Calculate the mass moment of inertia (I) required:
I
Để xem full tài liệu Xin vui long liên hệ page để được hỗ trợ
: https://www.facebook.com/thuvienluanvan01
HOẶC
https://www.facebook.com/garmentspace/
https://www.facebook.com/thuvienluanvan01
https://www.facebook.com/thuvienluanvan01
tai lieu tong hop, thu vien luan van, luan van tong hop, do an chuyen nganh
This document contains solutions to problems from Chapter 9 of a mechanical engineering design textbook. It analyzes welded joint designs and calculates shear stresses and forces. Key steps include selecting electrode materials based on member strengths, calculating primary and secondary shear stresses, and determining optimal weld leg sizes to satisfy a given allowable shear stress. Design specifications are provided for multiple welded joints.
The document provides solutions to example problems from a mechanics of materials textbook chapter. It includes:
1) Free body diagrams and calculations of internal forces and moments for determinate truss and frame structures.
2) Derivation of shear and moment diagrams by considering equilibrium of infinitesimal sections of beams.
3) Solutions involve solving systems of equations to determine reactions and internal forces.
The document contains 14 example problems solving for various values in gear design equations. Problem 14-1 solves for pressure angle, velocity, load, and bending stress. Problem 14-2 similarly solves for a different gear set. Problem 14-3 converts units and solves for velocity, load, and bending stress in MPa.
This document contains solved problems from Chapter 16 of an engineering textbook. Problem 16-1 involves calculating forces, pressures, and torques in a wet multi-disk brake system given various input parameters like shoe angles and dimensions. It finds the maximum pressure of 111.4 psi occurs on the right shoe for clockwise rotation. Problem 16-2 solves another wet multi-disk brake problem, showing a 2.7% reduction in torque is achieved using 25% less braking material. Problem 16-3 calculates reactions, pressures, and torques in a third wet multi-disk brake system using metric units and varying input parameters.
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
The document provides derivations of design equations for reinforced concrete beams. It begins by deriving the equation for maximum moment capacity of a singly reinforced beam based on concrete strength as M=0.167*fck*b*d^2. It then derives equations for doubly reinforced beams where compression steel is also required. The document further derives equations for design of flanged beams depending on whether the neutral axis lies within the flange or web. It concludes by outlining design procedures for singly and doubly reinforced beams.
Rolling Contact Bearing, Selection of Rolling Contact Bearings, Machine Element Design, Bantalan Gelinding, Pemilihan Bantalan Gelinding, Perancangan Elemen Mesin
Este documento resume los conceptos fundamentales de la corriente alterna trifásica. Explica cómo se genera mediante tres bobinados desfasados 120° entre sí y las configuraciones en estrella y triángulo. También analiza las cargas equilibradas y desequilibradas, calculando las tensiones, corrientes y potencias involucradas. Finalmente, incluye ejercicios numéricos para practicar los diferentes conceptos.
Este documento presenta una introducción a la ética y la deontología profesional. Define varios términos clave como "moral", "ética" y "deber", y discute diferentes perspectivas sobre lo moral, incluidas la moral como cumplimiento de deberes, la búsqueda de la felicidad y la moral de las virtudes comunitarias. El documento concluye que analizar estas perspectivas ayudará a los estudiantes a elegir y madurar sus propios criterios éticos para iluminar su realidad personal y profesional.
Marco legal del profecional en analista de sistemasParalafakyou Mens
1) El documento describe conceptos jurídicos básicos como el derecho, las fuentes del derecho (ley, costumbre, jurisprudencia y doctrina), las personas (físicas e ideales), la capacidad y responsabilidad. 2) Define a la persona como un ente capaz de adquirir derechos u obligaciones y distingue entre personas físicas e ideales. 3) Explica que las fuentes del derecho determinan las normas aplicables y las personas ideales (como empresas) tienen personalidad jurídica distinta a sus miembros.
Este documento proporciona instrucciones para crear, compilar y depurar un programa en COBOL usando Microfocus COBOL. Inicialmente se explica cómo abrir el entorno de desarrollo y crear un nuevo programa. Luego se detallan los pasos para compilar el código, ejecutarlo y depurarlo mediante la colocación de puntos de interrupción y la revisión de variables. Finalmente, se indica que este proceso de editar, compilar y ejecutar debe repetirse hasta que el programa funcione correctamente.
Este documento proporciona instrucciones para configurar el servidor de aplicaciones COBHTTPD. Explica cómo definir la información general como el puerto, el documento predeterminado y los directorios. También describe cómo configurar los proyectos y programas COBOL que se publicarán, y los compiladores compatibles. Proporciona detalles sobre cómo editar archivos XML y ejecutar el servidor y programas.
Este documento describe los elementos básicos del lenguaje de programación COBOL, incluyendo constantes figurativas como Zero y Space, constantes identificadas por nombre, identificadores, operadores aritméticos, de relación y lógicos, y cómo se evalúan las expresiones aritméticas y de BOOLE en COBOL.
Este documento presenta las instrucciones para un práctico de una asignatura de Problemática Política en la Universidad Nacional de Córdoba. El objetivo es que los estudiantes comprendan conceptos como democracia, regímenes políticos y neo liberalismo, y que puedan aplicarlos a la realidad latinoamericana. Como actividad, se les pide responder 4 preguntas relacionadas con los distintos tipos de regímenes democráticos en América Latina y los desafíos internos y externos para consolidar la democracia
La compañía de seguros necesita una base de datos para gestionar la información sobre los seguros que ofrece (hogar, vida y automóvil), los clientes y los agentes. La base de datos almacenará datos sobre los tipos de seguro, primas, clientes (nombre, dirección, etc.), agentes, beneficiarios y pólizas (fecha, detalles del seguro). Esto permitirá administrar las comisiones de los agentes y la información sobre los clientes y sus pólizas.
Este documento presenta la asignatura "Ética y deontología profesional" a los estudiantes. Explica que el objetivo es aclarar el significado de estos términos y justificar la necesidad de esta disciplina en la carrera. Resume las diferentes acepciones de términos como "moral", "ética" y "deontología" a lo largo de la historia. También describe brevemente diferentes enfoques de la moral como la búsqueda de la felicidad, el cumplimiento del deber, y la dialógica. El document
La guía explica cómo instalar ACUCOBOL en Windows 7 de 32 bits mediante la ejecución del archivo de instalación en modo de compatibilidad con Windows XP SP2 y seleccionando solo las suites de desarrollo durante la instalación. Adicionalmente, indica cómo verificar si el sistema es de 32 o 64 bits y sugiere usar una máquina virtual si el sistema es de 64 bits.
O documento fornece instruções para instalar e configurar o COBOL 4.5 no DOS, explicando como compilar e executar programas COBOL. Inclui detalhes sobre editar programas COBOL no DOS e no Windows.
This document provides information about the English for IT Level 1 course offered at Universidad Nacional de Córdoba in Argentina. It includes details such as the course validity period, classification as a complementary subject, weekly hours, and professors. The fundamentation section explains the importance of the course for developing the reading skills needed to access technical information in English. The general objectives are listed as acquiring reading comprehension abilities, vocabulary, and recognition of grammatical structures. The content is divided into 7 units covering topics such as basic reading comprehension techniques, sentence structures, verb tenses, and semantic fields. The teaching methodology involves both theoretical and practical components, with the gradual introduction of technical texts. Required and online references are also specified.
1. El documento presenta una introducción a los problemas de la ética normativa, incluyendo la fundamentación de normas morales, el origen de los principios morales y la aplicabilidad y rigurosidad de las normas. 2. Se describen posibles respuestas a estos problemas, como las fundamentaciones deontológicas y teleológicas, y posiciones como el heteronomismo, autonomismo, casuismo y situacionismo. 3. También se mencionan otros temas vinculados como la esencia de lo moral y problemas metafísicos como el libre albed
Este documento ofrece información sobre consideraciones para instalar y usar PowerCobol correctamente, así como sobre proyectos, programación, archivos, compilación, ejecución y el menú de PowerCobol. Explica cómo crear proyectos y ventanas, compilar y enlazar código, y ejecutar aplicaciones. También describe los objetos, propiedades y métodos que se usarán para programar interfaces gráficas en PowerCobol.
Este documento describe varios métodos para el montaje y desmontaje de rodamientos, incluyendo la inyección de aceite a presión, la dilatación térmica mediante calentamiento, la extracción por presión mecánica, y el montaje por impacto. También recomienda herramientas como llaves de gancho, martillo antirrebote y manguitos intermediarios para realizar estos procesos de manera segura.
Este documento proporciona una introducción a los conceptos básicos de la programación con Power Cobol, incluyendo proyectos, objetos, propiedades, métodos y eventos. Explica los diferentes tipos de objetos como etiquetas, cuadros de edición, botones y listas desplegables, y los eventos asociados a cada uno. También describe las secciones y declaraciones necesarias para crear una ventana y programar su comportamiento.
El documento describe las propiedades y usos del níquel y sus aleaciones. El níquel se utiliza comúnmente en aleaciones con cromo para formar aceros inoxidables, y con cobre para formar aleaciones como el Monel. Otras aleaciones notables son el Duraníquel, Permaníquel e Inconel, que combinan alta resistencia mecánica y resistencia a la corrosión para aplicaciones a alta temperatura.
Este documento describe las propiedades magnéticas de diferentes materiales. Explica que el magnetismo se produce por la interacción entre dipolos magnéticos y campos magnéticos externos. Algunos materiales como el hierro y el níquel son ferromagnéticos y pueden usarse en aplicaciones como generadores eléctricos y motores. La temperatura afecta el comportamiento magnético de los materiales.
Este documento proporciona información sobre los metales, en particular el aluminio. Resume que el aluminio es uno de los metales más utilizados debido a su bajo peso específico y propiedades mecánicas. Explica que el aluminio se obtiene principalmente de las bauxitas y se produce mediante electrolisis. También describe las propiedades, usos y aleaciones más comunes del aluminio.
1. Chapter 11
11-1 For the deep-groove 02-series ball bearing with R = 0.90, the design life xD, in multiples
of rating life, is
xD = 30 000(300)(60)
106
= 540 Ans.
The design radial load FD is
FD = 1.2(1.898) = 2.278 kN
From Eq. (11-6),
C10 = 2.278
540
0.02 + 4.439[ln(1/0.9)]1/1.483
1/3
= 18.59 kN Ans.
Table 11-2: Choose a 02-30 mm with C10 = 19.5 kN. Ans.
Eq. (11-18):
R = exp
−
540(2.278/19.5)3 − 0.02
4.439
1.483
= 0.919 Ans.
11-2 For the Angular-contact 02-series ball bearing as described, the rating life multiple is
xD = 50 000(480)(60)
106
= 1440
The design load is radial and equal to
FD = 1.4(610) = 854 lbf = 3.80 kN
Eq. (11-6):
C10 = 854
1440
0.02 + 4.439[ln(1/0.9)]1/1.483
1/3
= 9665 lbf = 43.0 kN
Table 11-2: Select a 02-55 mm with C10 = 46.2 kN. Ans.
Using Eq. (11-18),
R = exp
−
1440(3.8/46.2)3 − 0.02
4.439
1.483
= 0.927 Ans.
2. 298 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
11-3 For the straight-Roller 03-series bearing selection, xD = 1440 rating lives from Prob. 11-2
solution.
FD = 1.4(1650) = 2310 lbf = 10.279 kN
C10 = 10.279
1440
1
3/10
= 91.1 kN
Table 11-3: Select a 03-55 mm with C10 = 102 kN. Ans.
Using Eq. (11-18),
R = exp
−
1440(10.28/102)10/3 − 0.02
4.439
1.483
= 0.942 Ans.
11-4 We can choose a reliability goal of
√
0.90 = 0.95 for each bearing. We make the selec-tions,
find the existing reliabilities, multiply them together, and observe that the reliability
goal is exceeded due to the roundup of capacity upon table entry.
Another possibility is to use the reliability of one bearing, say R1. Then set the relia-bility
goal of the second as
R2 = 0.90
R1
or vice versa. This gives three pairs of selections to compare in terms of cost, geometry im-plications,
etc.
11-5 Establish a reliability goal of
√
0.90 = 0.95 for each bearing. For a 02-series angular con-tact
ball bearing,
C10 = 854
1440
0.02 + 4.439[ln(1/0.95)]1/1.483
1/3
= 11 315 lbf = 50.4 kN
Select a 02-60 mm angular-contact bearing with C10 = 55.9 kN.
RA = exp
−
1440(3.8/55.9)3 − 0.02
4.439
1.483
= 0.969
For a 03-series straight-roller bearing,
C10 = 10.279
1440
0.02 + 4.439[ln(1/0.95)]1/1.483
3/10
= 105.2 kN
Select a 03-60 mm straight-roller bearing with C10 = 123 kN.
RB = exp
−
1440(10.28/123)10/3 − 0.02
4.439
1.483
= 0.977
3. Chapter 11 299
Form a table of existing reliabilities
Rgoal RA RB 0.912
0.90 0.927 0.941 0.872
0.95 0.969 0.977 0.947
0.906
The possible products in the body of the table are displayed to the right of the table. One,
0.872, is predictably less than the overall reliability goal. The remaining three are the
choices for a combined reliability goal of 0.90. Choices can be compared for the cost of
bearings, outside diameter considerations, bore implications for shaft modifications and
housing modifications.
The point is that the designer has choices. Discover them before making the selection
decision. Did the answer to Prob. 11-4 uncover the possibilities?
To reduce the work to fill in the body of the table above, a computer program can be
helpful.
11-6 Choose a 02-series ball bearing from manufacturer #2, having a service factor of 1. For
Fr = 8 kN and Fa = 4 kN
xD = 5000(900)(60)
106
= 270
Eq. (11-5):
C10 = 8
270
0.02 + 4.439[ln(1/0.90)]1/1.483
1/3
= 51.8 kN
Trial #1: From Table (11-2) make a tentative selection of a deep-groove 02-70 mm with
C0 = 37.5 kN.
Fa
C0
= 4
37.5
= 0.107
Table 11-1:
Fa/(V Fr ) = 0.5 e
X2 = 0.56, Y2 = 1.46
Eq. (11-9):
Fe = 0.56(1)(8) + 1.46(4) = 10.32 kN
Eq. (11-6):
C10 = 10.32
270
1
1/3
= 66.7 kN 61.8 kN
Trial #2: From Table 11-2 choose a 02-80 mm having C10 = 70.2 and C0 = 45.0.
Check:
Fa
C0
= 4
45
= 0.089
4. 300 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Table 11-1: X2 = 0.56, Y2 = 1.53
Fe = 0.56(8) + 1.53(4) = 10.60 kN
Eq. (11-6):
C10 = 10.60
270
1
1/3
= 68.51 kN 70.2 kN
∴ Selection stands.
Decision: Specify a 02-80 mm deep-groove ball bearing. Ans.
11-7 From Prob. 11-6, xD = 270 and the final value of Fe is 10.60 kN.
C10 = 10.6
270
0.02 + 4.439[ln(1/0.96)]1/1.483
1/3
= 84.47 kN
Table 11-2: Choose a deep-groove ball bearing, based upon C10 load ratings.
Trial #1:
Tentatively select a 02-90 mm.
C10 = 95.6, C0 = 62 kN
Fa
4
= = 0.0645
C0
62
From Table 11-1, interpolate for Y2.
Fa/C0 Y2
0.056 1.71
0.0645 Y2
0.070 1.63
Y2 − 1.71
1.63 − 1.71
= 0.0645 − 0.056
0.070 − 0.056
= 0.607
Y2 = 1.71 + 0.607(1.63 − 1.71) = 1.661
Fe = 0.56(8) + 1.661(4) = 11.12 kN
C10 = 11.12
270
0.02 + 4.439[ln(1/0.96)]1/1.483
1/3
= 88.61 kN 95.6 kN
Bearing is OK.
Decision: Specify a deep-groove 02-90 mm ball bearing. Ans.
5. Chapter 11 301
11-8 For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.90 and
Fr = 12 kN
xD = 4000(750)(60)
106
= 180
C10 = 12
180
1
3/10
= 57.0 kN Ans.
11-9
Assume concentrated forces as shown.
Pz = 8(24) = 192 lbf
Py = 8(30) = 240 lbf
T = 192(2) = 384 lbf · in
T x = −384 + 1.5F cos 20◦ = 0
F = 384
1.5(0.940)
= 272 lbf
MzO
= 5.75Py + 11.5Ry
A
− 14.25F sin 20◦ = 0;
thus 5.75(240) + 11.5Ry
A
− 14.25(272)(0.342) = 0
Ry
A
= −4.73 lbf
My
O
= −5.75Pz − 11.5Rz
A
− 14.25F cos 20◦ = 0;
thus −5.75(192) − 11.5Rz
A
− 14.25(272)(0.940) = 0
Rz
A
= −413 lbf; RA = [(−413)2 + (−4.73)2]1/2 = 413 lbf
Fz = RzO
+ Pz + Rz
A
+ F cos 20◦ = 0
RzO
+ 192 − 413 + 272(0.940) = 0
RzO
= −34.7 lbf
B
O
z
11
1
2
Rz
O
Ry
O
Pz
Py
T
F
20
Ry
A
Rz
A
A
T
y
2
3
4
x
6. 302 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Fy = Ry
O
+ Py + Ry
A
− F sin 20◦ = 0
Ry
O
+ 240 − 4.73 − 272(0.342) = 0
Ry
O
= −142 lbf
RO = [(−34.6)2 + (−142)2]1/2 = 146 lbf
So the reaction at A governs.
Reliability Goal:
√
0.92 = 0.96
FD = 1.2(413) = 496 lbf
xD = 30 000(300)(60/106) = 540
540
C10 = 496
0.02 + 4.439[ln(1/0.96)]1/1.483
1/3
= 4980 lbf = 22.16 kN
A 02-35 bearing will do.
Decision: Specify an angular-contact 02-35 mm ball bearing for the locations at A and O.
Check combined reliability. Ans.
11-10 For a combined reliability goal of 0.90, use
√
0.90 = 0.95 for the individual bearings.
x0 = 50 000(480)(60)
106
= 1440
z
O
20
A
16
10
FA
y
RO
RB B
C
FC
x
20
The resultant of the given forces are RO = 607 lbf and RB = 1646 lbf.
At O: Fe = 1.4(607) = 850 lbf
Ball: C10 = 850
1440
0.02 + 4.439[ln(1/0.95)]1/1.483
1/3
= 11 262 lbf or 50.1 kN
Select a 02-60 mm angular-contact ball bearing with a basic load rating of 55.9 kN.
At B: Fe = 1.4(1646) = 2304 lbf
Roller: C10 = 2304
1440
0.02 + 4.439[ln(1/0.95)]1/1.483
3/10
= 23 576 lbf or 104.9 kN
Select a 02-80 mm cylindrical roller or a 03-60 mm cylindrical roller. The 03-series roller
has the same bore as the 02-series ball.
7. Chapter 11 303
11-11 The reliability of the individual bearings is R =
√
0.999 = 0.9995
From statics,
Ry
O
y
Ry
O
O
300
A
Fy
A
400
Fz
A
Rz
O
z
= −163.4 N, RzO
150
Ry
E
E
C
Rz
E
FC
x
= 107 N, RO = 195 N
Ry
E
= −89.1 N, RzE
= −174.4 N, RE = 196 N
xD = 60 000(1200)(60)
106
= 4320
C10 = 0.196
4340
0.02 + 4.439[ln(1/0.9995)]1/1.483
1/3
= 8.9 kN
A 02-25 mm deep-groove ball bearing has a basic load rating of 14.0 kN which is ample.
An extra-light bearing could also be investigated.
11-12 Given:
Fr A = 560 lbf or 2.492 kN
Fr B = 1095 lbf or 4.873 kN
Trial #1: Use KA = KB = 1.5 and from Table 11-6 choose an indirect mounting.
0.47Fr A
KA
?
0.47Fr B
KB
− (−1)(0)
0.47(2.492)
1.5
?
0.47(4.873)
1.5
0.781 1.527 Therefore use the upper line of Table 11-6.
FaA = FaB = 0.47Fr B
KB
= 1.527 kN
PA = 0.4Fr A + KAFaA = 0.4(2.492) + 1.5(1.527) = 3.29 kN
PB = Fr B = 4.873 kN
8. 304 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Fig. 11-16: fT = 0.8
Fig. 11-17: fV = 1.07
Thus, a3l = fT fV = 0.8(1.07) = 0.856
√
Individual reliability: Ri =
0.9 = 0.95
Eq. (11-17):
(C10)A = 1.4(3.29)
40 000(400)(60)
4.48(0.856)(1 − 0.95)2/3(90)(106)
0.3
= 11.40 kN
(C10)B = 1.4(4.873)
40 000(400)(60)
4.48(0.856)(1 − 0.95)2/3(90)(106)
0.3
= 16.88 kN
From Fig. 11-14, choose cone 32 305 and cup 32 305 which provide Fr = 17.4 kN and
K = 1.95. With K = 1.95 for both bearings, a second trial validates the choice of cone
32 305 and cup 32 305. Ans.
11-13
R =
√
0.95 = 0.975
T = 240(12)(cos 20◦) = 2706 lbf · in
F = 2706
6 cos 25◦
= 498 lbf
In xy-plane:
MO = −82.1(16) − 210(30) + 42Ry
C
= 0
Ry
C
= 181 lbf
Ry
O
= 82 + 210 − 181 = 111 lbf
In xz-plane:
MO = 226(16) − 452(30) − 42Rz
c
= 0
RzC
= −237 lbf
RzO
= 226 − 451 + 237 = 12 lbf
RO = (1112 + 122)1/2 = 112 lbf Ans.
RC = (1812 + 2372)1/2 = 298 lbf Ans.
FeO = 1.2(112) = 134.4 lbf
FeC = 1.2(298) = 357.6 lbf
xD = 40 000(200)(60)
106
= 480
z
14
16
12
Rz
O
Rz
C
Ry
O
A
B
C
Ry
C
O
451
210
226
T
T
82.1
x
y
9. Chapter 11 305
(C10)O = 134.4
480
0.02 + 4.439[ln(1/0.975)]1/1.483
1/3
= 1438 lbf or 6.398 kN
(C10)C = 357.6
480
0.02 + 4.439[ln(1/0.975)]1/1.483
1/3
= 3825 lbf or 17.02 kN
Bearing at O: Choose a deep-groove 02-12 mm. Ans.
Bearing at C: Choose a deep-groove 02-30 mm. Ans.
There may be an advantage to the identical 02-30 mm bearings in a gear-reduction unit.
11-14 Shafts subjected to thrust can be constrained by bearings, one of which supports the thrust.
The shaft floats within the endplay of the second (Roller) bearing. Since the thrust force
here is larger than any radial load, the bearing absorbing the thrust is heavily loaded com-pared
to the other bearing. The second bearing is thus oversized and does not contribute
measurably to the chance of failure. This is predictable. The reliability goal is not
√
0.99,
but 0.99 for the ball bearing. The reliability of the roller is 1. Beginning here saves effort.
Bearing at A (Ball)
Fr = (362 + 2122)1/2 = 215 lbf = 0.957 kN
Fa = 555 lbf = 2.47 kN
Trial #1:
Tentatively select a 02-85 mm angular-contact with C10 = 90.4 kN and C0 = 63.0 kN.
Fa
C0
= 2.47
63.0
= 0.0392
xD = 25 000(600)(60)
106
= 900
Table 11-1: X2 = 0.56, Y2 = 1.88
Fe = 0.56(0.957) + 1.88(2.47) = 5.18 kN
FD = fAFe = 1.3(5.18) = 6.73 kN
900
C10 = 6.73
0.02 + 4.439[ln(1/0.99)]1/1.483
1/3
= 107.7 kN 90.4 kN
Trial #2:
Tentatively select a 02-95 mm angular-contact ball with C10 = 121 kN and C0 = 85 kN.
Fa
C0
= 2.47
85
= 0.029
10. 306 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Table 11-1: Y2 = 1.98
Fe = 0.56(0.957) + 1.98(2.47) = 5.43 kN
FD = 1.3(5.43) = 7.05 kN
900
C10 = 7.05
0.02 + 4.439[ln(1/0.99)]1/1.483
1/3
= 113 kN 121 kN O.K.
Select a 02-95 mm angular-contact ball bearing. Ans.
Bearing at B (Roller): Any bearing will do since R = 1. Let’s prove it. From Eq. (11-18)
when
af FD
C10
3
xD x0 R = 1
The smallest 02-series roller has a C10 = 16.8 kN for a basic load rating.
0.427
16.8
3
(900) ? 0.02
0.0148 0.02 ∴ R = 1
Spotting this early avoided rework from
√
0.99 = 0.995.
Any 02-series roller bearing will do. Same bore or outside diameter is a common choice.
(Why?) Ans.
11-15 Hoover Ball-bearing Division uses the same 2-parameter Weibull model as Timken:
b = 1.5, θ = 4.48. We have some data. Let’s estimate parameters b and θ from it. In
Fig. 11-5, we will use line AB. In this case, B is to the right of A.
For F = 18 kN, (x)1 = 115(2000)(16)
106
= 13.8
This establishes point 1 on the R = 0.90 line.
1
0
2
100
18
10
0 1
1 2
39.6
1
10
13.8 72
1
100 x
2 log x
F
A B
log F
R 0.90
R 0.20
11. Chapter 11 307
The R = 0.20 locus is above and parallel to the R = 0.90 locus. For the two-parameter
Weibull distribution, x0 = 0 and points A and B are related by:
xA = θ[ln(1/0.90)]1/b (1)
xB = θ[ln(1/0.20)]1/b
and xB/xA is in the same ratio as 600/115. Eliminating θ
b = ln[ln(1/0.20)/ ln(1/0.90)]
ln(600/115)
= 1.65
Solving for θ in Eq. (1)
θ = xA
[ln(1/RA)]1/1.65
= 1
[ln(1/0.90)]1/1.65
= 3.91
Therefore, for the data at hand,
R = exp
−
x
3.91
1.65
Check R at point B: xB = (600/115) = 5.217
R = exp
−
5.217
3.91
1.65
= 0.20
Note also, for point 2 on the R = 0.20 line.
log(5.217) − log(1) = log(xm)2 − log(13.8)
(xm)2 = 72
11-16 This problem is rich in useful variations. Here is one.
Decision: Make straight roller bearings identical on a given shaft. Use a reliability goal of
(0.99)1/6 = 0.9983.
Shaft a
FrA
= (2392 + 1112)1/2 = 264 lbf or 1.175 kN
FrB
= (5022 + 10752)1/2 = 1186 lbf or 5.28 kN
Thus the bearing at B controls
xD = 10 000(1200)(60)
106
= 720
0.02 + 4.439[ln(1/0.9983)]1/1.483 = 0.080 26
C10 = 1.2(5.2)
720
0.080 26
0.3
= 97.2 kN
Select either a 02-80 mm with C10 = 106 kN or a 03-55 mm with C10 = 102 kN
12. 308 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Shaft b
FrC
= (8742 + 22742)1/2 = 2436 lbf or 10.84 kN
FrD
= (3932 + 6572)1/2 = 766 lbf or 3.41 kN
The bearing at C controls
xD = 10 000(240)(60)
106
= 144
C10 = 1.2(10.84)
144
0.0826
0.3
= 122 kN
Select either a 02-90 mm with C10 = 142 kN or a 03-60 mm with C10 = 123 kN
Shaft c
FrE
= (11132 + 23852)1/2 = 2632 lbf or 11.71 kN
FrF
= (4172 + 8952)1/2 = 987 lbf or 4.39 kN
The bearing at E controls
xD = 10 000(80)(60/106) = 48
C10 = 1.2(11.71)
48
0.0826
0.3
= 94.8 kN
Select a 02-80 mm with C10 = 106 kN or a 03-60 mm with C10 = 123 kN
11-17 The horizontal separation of the R = 0.90 loci in a log F-log x plot such as Fig. 11-5
will be demonstrated. We refer to the solution of Prob. 11-15 to plot point G (F =
18 kN, xG = 13.8). We know that (C10)1 = 39.6 kN, x1 = 1. This establishes the unim-proved
steel R = 0.90 locus, line AG. For the improved steel
(xm)1 = 360(2000)(60)
106
= 43.2
We plot point G(F = 18 kN, xG = 43.2), and draw the R = 0.90 locus AmG parallel
to AG
Am
1
0
2
100
55.8
18
10
0 1
G G
39.6
1
10
13.8
1
100
2
x
log x
F
A
Improved steel
log F
Unimproved steel
43.2
R 0.90
R 0.90
1
3
1
3
13. Chapter 11 309
We can calculate (C10)m by similar triangles.
log(C10)m − log 18
log 43.2 − log 1
= log 39.6 − log 18
log 13.8 − log 1
log(C10)m = log 43.2
log 13.8
log
39.6
18
+ log 18
(C10)m = 55.8 kN
The usefulness of this plot is evident. The improvement is 43.2/13.8 = 3.13 fold in life.
This result is also available by (L10)m/(L10)1 as 360/115 or 3.13 fold, but the plot shows
the improvement is for all loading. Thus, the manufacturer’s assertion that there is at least
a 3-fold increase in life has been demonstrated by the sample data given. Ans.
11-18 Express Eq. (11-1) as
Fa
1 L1 = Ca
10L10 = K
For a ball bearing, a = 3 and for a 02-30 mm angular contact bearing, C10 = 20.3 kN.
K = (20.3)3(106) = 8.365(109)
At a load of 18 kN, life L1 is given by:
L1 = K
Fa
1
= 8.365(109)
183
= 1.434(106) rev
For a load of 30 kN, life L2 is:
L2 = 8.365(109)
303
= 0.310(106) rev
In this case, Eq. (7-57) – the Palmgren-Miner cycle ratio summation rule – can be ex-pressed
as
l1
L1
+ l2
L2
= 1
Substituting,
200 000
1.434(106)
+ l2
0.310(106)
= 1
l2 = 0.267(106) rev Ans.
Check:
200 000
1.434(106)
+ 0.267(106)
0.310(106)
= 1 O.K.
11-19 Total life in revolutions
Let:
l = total turns
f1 = fraction of turns at F1
f2 = fraction of turns at F2
14. 310 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
From the solution of Prob. 11-18, L1 = 1.434(106) rev and L2 = 0.310(106) rev.
Palmgren-Miner rule:
l1
L1
+ l2
L2
= f1l
L1
+ f2l
L2
= 1
from which
l = 1
f1/L1 + f2/L2
l = 1
{0.40/[1.434(106)]} + {0.60/[0.310(106)]}
= 451 585 rev Ans.
Total life in loading cycles
4 min at 2000 rev/min = 8000 rev
6 min
10 min/cycle
at 2000 rev/min = 12 000 rev
20 000 rev/cycle
451 585 rev
20 000 rev/cycle
= 22.58 cycles Ans.
Total life in hours
10
min
cycle
22.58 cycles
60 min/h
= 3.76 h Ans.
11-20 While we made some use of the log F-log x plot in Probs. 11-15 and 11-17, the principal
use of Fig. 11-5 is to understand equations (11-6) and (11-7) in the discovery of the cata-log
basic load rating for a case at hand.
Point D
FD = 495.6 lbf
log FD = log 495.6 = 2.70
xD = 30 000(300)(60)
106
= 540
log xD = log 540 = 2.73
KD = F3D
xD = (495.6)3(540)
= 65.7(109) lbf3 · turns
log KD = log[65.7(109)] = 10.82
FD has the following uses: Fdesign, Fdesired, Fe when a thrust load is present. It can include
application factor af , or not. It depends on context.
15. Chapter 11 311
Point B
xB = 0.02 + 4.439[ln(1/0.99)]1/1.483
= 0.220 turns
log xB = log 0.220 = −0.658
FB = FD
xD
xB
1/3
= 495.6
540
0.220
1/3
= 6685 lbf
Note: Example 11-3 used Eq. (11-7). Whereas, here we basically used Eq. (11-6).
log FB = log(6685) = 3.825
KD = 66853(0.220) = 65.7(109) lbf3 · turns (as it should)
Point A
FA = FB = C10 = 6685 lbf
logC10 = log(6685) = 3.825
xA = 1
log xA = log(1) = 0
K10 = F3A
xA = C3
10(1) = 66853 = 299(109) lbf3 · turns
Note that KD/K10 = 65.7(109)/[299(109)] = 0.220, which is xB. This is worth knowing
since
K10 = KD
xB
log K10 = log[299(109)] = 11.48
0.658
B
0.1
1
A
1
0
10
1
102
2
F
log F
4 104
6685
103
495.6
3
2 102
D
103
3
x
log x
540
Now C10 = 6685 lbf = 29.748 kN, which is required for a reliability goal of 0.99. If we
select an angular contact 02-40 mm ball bearing, then C10 = 31.9 kN = 7169 lbf.