Projetos e Elementos de Maquinas - Juvinall -exercicios Resolvidos - Mecheng Broad Perspective

SOLUTION (1.1D)\nKnown: Definitions of the words science, engineering, and design are given in Section 1.1.\n\nFind: Write definitions of above words using a dictionary and compare with those given in Section 1.1.\n\nAnalysis:\n1. According to Webster's Ninth New Collegiate Dictionary, science is the application of science and mathematics by which the properties of matter and the sources of energy in nature are made useful to people in structures, machines, products, systems, and processes. According to Section 1.1, engineering is an applied science, in the sense that it is concerned with understanding scientific principles and applying them to achieve a designated goal utilizing the resources and laws of nature to benefit humanity.\n\n2. According to Webster's Ninth New Collegiate Dictionary, art is a skill acquired by experience, study, or observation. (Also, art is defined as the conscious use of skill and creative imagination, especially in the production of aesthetic objects.)\n\n3. According to Webster's Ninth New Collegiate Dictionary, design is an act of devising for a specific function or end. According to Section 1.1, mechanical engineering design deals with the conception, design, development, refinement, and application of machines and mechanical apparatus of all kinds.\n\nSOLUTION (1.2D)\nKnown: Safety devices are found in machines used in automatic production operations. These safety devices stop the machine when a faulty operation takes place.\n\nFind: Describe and sketch several mechanical safety devices.\n\nAnalysis: An initial search of the patent literature at http://patent.womplex.ibm. com for mechanical safety devices yielded (1) a chain saw cutting chain safety stop and (2) a safety stop switch for the control unit for a marine engine:\n\n(1) US4651423: Chain saw cutting chain safety stop\nDisclosed is a safety stop that is operatively connected to the existing rotatively mounted front handle of a chain saw to independently actuate such safety stop to its actuated position to stop the rotation of the cutting chain upon rotation of such rotatively mounted front handle either automatically when \"kick-back\" occurs or upon such rotation by the operator at his option.\n\n1-1 (2) US4250358: Control unit for marine engines employing safety stop switch\nA safety stop switch for a marine engine includes a control unit housing, and a normally closed, single throw toggle switch mounted on the housing and including a switch arm extending away from the housing to permit operation of the switch between closed and open positions. The switch is electrically connected to the engine to interrupt engine operation when switched to the open position. A hood extends from the periphery of the housing and covers the switch arm when in the closed position, and has a peripheral lip which is spaced from the switch arm a dimension which is less than the thickness of a key which encircles the switch arm. A lanyard is connected to the key and to the boat operator, and upon the exertion of a force on the lanyard, the key pulls the switch arm down into the open position, thereby interrupting operation of the marine engine.\n\n1-2 Comments:\n1. With additional effort, safety devices for automatic production operation should be found.\n2. For additional safety devices see Mechanical Details for Product Design, Douglas C. Greenwood, editor, McGraw-Hill Book Company, New York, 1964.\n\n1-3 SOLUTION (1.3D)\nKnown: The OSHA regulations are found at http://www.osha.gov.\nFind: List methods used to guard machine hazards. Give conditions where guards\nshould be used.\nAnalysis: The Code of Federal Regulations, Title 29--Labor, § 1910.212 lists general\nrequirements for all machines:\n\n(a) Machine guarding--(1) Types of guarding. One or more methods of machine\nguarding shall be provided to protect the operator and other employees in the machine\narea from hazards such as those created by point of operation, ingoing nip points,\nrotating parts, flying chips and sparks. Examples of guarding methods are-barrier\nguards, two-hand tripping devices,electronic safety devices, etc.\n\n(2) General requirements for machine guards. Guards shall be affixed to the\nmachine where possible and secured elsewhere if for any reason attachment to the\nmachine is not possible. The guard shall be such that it does not offer an accident\nhazard in itself.\n\n(3) Point of operation guarding. (i) Point of operation is the area on a machine\nwhere work is actually performed upon the material being processed.\n(ii) The point of operation of machines whose operation exposes an employee to\ninjury, shall be guarded. The guarding device shall be in conformity with any\nappropriate standards therefor, or, in the absence of applicable specific standards, shall\nbe so designed and constructed as to prevent the operator from having any part of his\nbody in the danger zone during the operating cycle.\n(iii) Special hand tools for placing and removing material shall be such as to\npermit easy handling of material without the operator placing a hand in the danger\nzone. Such tools shall not be used in lieu of other guarding required by this section, but can\nbe used to supplement protection provided.\n(iv) The following are some of the machines which usually require point of\noperation guarding:\n(a) Guillotine cutters.\n(b) Shears.\n(c) Alligator shears.\n(d) Power presses.\n(e) Milling machines.\n(f) Power saws.\n(g) Jointers.\n(h) Portable power tools.\n(i) Forming rolls and calenders.\n\n(4) Barrels, containers, and drums. Revolving drums, barrels, and containers\nshall be guarded by an enclosure which is interlocked with the drive mechanism, so\nthat the barrel, drum, or container cannot revolve unless the guard enclosure is in\nplace.\n\n(5) Exposure of blades. When the periphery of the blades of a fan is less than\nseven (7) feet above the floor or working level, the blades shall be guarded. The guard\nshall have openings no larger than one-half (1/2) inch.\n\n(b) Anchoring fixed machinery. Machines designed for a fixed location shall be\nsecurely anchored to prevent walking or moving. SOLUTION (1.4D)\nKnown: The web site http://www.nssn.org lists titles and organizations for standards.\nFind: List titles and organizations for standards on (a) machine guarding, (b) refuse vehicles, and (c) portable grinders.\nAnalysis:\n\nTABLE OF ORGANIZATIONS AND STANDARDS\nOrganization Standards\n(a) machine guarding U.S. Government\n(OSHA) Code of Federal\n Regulations, Title 29--\n Labor, § 1910.212\n(b) refuse vehicles American National\nStandards Institute ANSI Z245.1\n(c) portable grinders American National\nStandards Institute ANSI B7.1 (abrasive\n wheels)\n U.S. Government\n (OSHA) Code of Federal\n Regulations, §1926.303\n (abrasive wheels and\ntools)\n\nSOLUTION (1.12D)\nKnown: An essay must be written.\nFind: Write an essay titled \"What Makes a Successful Design Engineer?\"\nAnalysis:\n1. This exercise is left to the student.\n2. For comparison, an article by George H. Logan published in 1959, in Product\nEngineering Design Manual, edited by Douglas C. Greenwood, McGraw-Hill\nBook Company, New York, p. x suggests the following:\n\nA design engineer must have the inner spark that urges him to obtain the inner\nsatisfaction linked with achievement. The designer should look for unorthodox\napproaches to problems. The design engineer should be able to control his mind\nto allow for clear and unobstructed thinking. The designer should believe in and\npractice work organization. An engineer who lacks organization can waste much\ntime through imperfect understanding of what the project is supposed to produce.\nThe engineer should discipline his mind, control his speech and actions so that he\nachieves full emotional balance. Building a calm temperament is not always\neasy, but it can be done if restraint is practiced. The importance of broadening\nthe knowledge of non-technical business topics should not be overlooked. The\nengineer must be able to make quick decisions that are correct most of the time\nand the creative design engineer can make for himself an opportunity that leads to\nan outstanding position in his chosen field. SOLUTION (1.14)\nKnown: The mass and the acceleration of gravity are given.\nFind: Determine the weight of the object in:\n(a) English Engineering units\n(b) British Gravitational units\n(c) SI units\nSchematic and Given Data:\n\n10 kg\n9.81 m/s²\n\nAnalysis:\n(a) English Engineering units\nF = ma/gc [Eq. (1.1a)]\nwhere\nm = (10 kg / 1 slug) = 0.685 slug\nor,\n\nm = 0.685 slug (32.2 lbm / 1 slug) = 22.1 lbm\n\n a = (9.81 m/s²) (1 ft / 0.3048 m) = 32.2 ft/s²\n\ngc = 32.2 ft-lbm/lb s²\n\nThus,\nF = (22.057 lbm) (32.2 ft/s²) = 22.1 lb\n\n(b) British Gravitational units\nF = ma [Eq. (1.1b)]\n\nF = (0.685 slug)(32.2 ft/s²) = 22.1 slug-ft/s² = 22.1 lb\n\n(c) SI units\nF = ma [Eq. (1.1c)]\nF = (10 kg)(9.81 m/s²) = 98.1 kg·m/s² = 98.1 N\n\nComment: The answers in (a) and (b) are equivalent to the answer in (c), since\n(98.1 N)(1 lb) / (4.448 N) = 22.1 lb. SOLUTION (1.15)\nKnown: The parameters m, a, F, W, s, ω, T and \\(\\dot{W}\\) are identified.\n\nFind: Check the dimensional homogeneity of the following equations: (a) F = ma, (b) W = Fs, (c) \\(\\dot{W} = Tω\\).\n\nGiven Data:\n m = mass\na = acceleration\nF = force\nW = work\ns = distance\nω = angular velocity\nT = torque\n\\(\\dot{W} = power\\)\n\nAnalysis:\n1. Let the dimensions of length, mass, and time be given by\n length [L]\nmass [M]\ntime [T]\n \nThen, m [M]\na [L/T²]\nF [M L/T²]\nW [M L²/T²]\ns [L]\nω [1/T]\nT [M L²/T²]\n\\(\\dot{W} [M L²/T³]\\)\n\n2. (a) F = ma\nML/T² = M(L/T²) = ML/T²\n(b) W = Fs\nML²/T² = (ML/T²)(L) = ML²/T²\n(c) \\(\\dot{W} = Tω\n(ML²/T²)(1/T) = ML²/T³\n SOLUTION (1.16)\nKnown: An object has known mass and volume.\n\nFind:\n(a) Determine the object weight and average density at a location where g = 9.55 m/s².\n(b) Determine the object weight and average density at a location where g = 1.7 m/s².\n\nSchematic and Given Data:\n m = 7.8 kg\n V = 0.7 m³\ng = 9.55 m/s²\n g = 1.7 m/s²\n(a) On earth\n(b) On moon\n\nAnalysis:\n(a) On the earth where g = 9.55 m/s², Weight = F = ma [Eq. (1.1c)]\nF = (7.8 kg)(9.55 m/s²) = 74.5 N\nAverage density, ρ = m/V\nρ = (7.8 kg/0.7 m³) = 11.1 kg/m³\n\n(b) On the moon where g = 1.7 m/s², Weight = F = ma.\nF = (7.8 kg)(1.7 m/s²) = 13.3 N\nρ = (7.8 kg/(0.7 m³) = 11.1 kg/m³\n\nComment: The weight is dependent on gravity while the density is independent of gravity.\n SOLUTION (1.17)\nKnown: A spacecraft component with known volume and weight is located where the acceleration of gravity is 31.0 ft/s².\n\nFind: Determine its weight and its average density on the moon, where g = 5.57 ft/s².\n\nSchematic and Given Data:\n V = 8 ft³\n g = 31.0 ft/s²\n g = 5.57 ft/s²\n25 lb\n\nAnalysis:\n1. Using British Gravitational units, and F = ma [Eq. (1.1b)]\n m = F/g = (25 lb)/(31.0 ft/s²) = 0.806 slug\n2. On the moon:\n F = (0.806 slug)(5.57 ft/s²) = 4.49 lb\n ρ = (0.806)/(32.2) lbm/8 ft³ = 3.24 lbm/ft³\nSOLUTION (1.18)\nKnown: An object is suspended from a spring at a location where g = 9.81 m/s². The deflection of the spring is known.\n\nFind: Determine the mass of the object.\n\nSchematic and Given Data:\n g = 9.81 m/s²\n\nAssumption: The spring has a linear force-deflection curve.\n\nAnalysis:\n1. The weight of the object is (30 mm)(1 N/5 mm) = 6 N\n2. Using Eq. (1.1c), F = ma\n m = F/a = 6 N/9.81 m/s² = 0.612 kg\n SOLUTION (1.19) Known: The British Gravitational System uses the mass unit slug. By definition, a mass of 1 slug is accelerated at a rate of 1 ft/s2 by a force of 1 lb.\n\nFind: Explain why the slug is a convenient mass unit.\n\nAnalysis: Unlike the English Engineering System, the British Gravitational System eliminates the need for gc in F = ma since gc = 1.\n\nSOLUTION (1.20) Known: An object with known weight is located where the acceleration of gravity is g = 30.5 ft/s2.\n\nFind: Determine the magnitude of the net force required to accelerate the object at 25 ft/s2.\n\nSchematic and Given Data:\n1. a = 25 ft/s2\n 20 lb\n g = 30.5 ft/s2\n\nAnalysis:\n1. Using British Gravitational units, F = ma [Eq. (1.1b)]\n m = F/g = 20 lb / 30.5 ft/s2 = 0.656 slug\n2. F = ma = (0.656 slug)(25 ft/s2) = 16.4 lb.\n\nSOLUTION (1.21) Known: An automobile passenger experiences a deceleration of 50 g’s in a head-on crash.\n\nFind: Determine the force experienced by the automobile passenger.\n\n1-10