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牙刷自动直线式平磨毛一体机设计
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牙刷自动直线式平磨毛一体机设计 (2),牙刷自动直线式平磨毛一体机设计,(2),牙刷,自动,直线,式平磨毛,一体机,设计
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广陵学院毕业设计外文资料翻译 学 生 姓 名： 仇 洋 洋 学 号： 100007106 专 业： 机械设计制造及其自动化 设 计 题 目： 牙刷自动直线式平磨毛一体机设计 指 导 老 师： 竺志大 翻 译 文 献： An Introduction to Mechanical Engineering 文 献 作 者： Jonathan Wickert 出 版 社： 西安交通大学出版社 出 版 时 间： 2003-10-1 版 2014年 3月 27日5.5 FACTOR OF SAFETY Mechanical engineers determine the shape, dimensions, and materials for a wide range of hardware. The analyses that support those design decisions take into account the tensile, compressive, and shear stresses that are present. Designers are aware that a component can break or otherwise be rendered useless through a variety of mechanisms. It could yield and take on a permanently deformed shape, fracture suddenly into many pieces, or be damaged through corrosion. Mechanical engineers perform the calculations and tests that are necessary to ensure reliability against those failure modes and other possible ones. Engineers rely on analysis, experiments, judgment, and design codes when they make decisions of that nature. In this section, we discuss failure analysis for the prototypical case of yielding in either tension or shear. Such an analysis predicts the onset of yielding in ductile stress. Yielding is useful to prevent the materials from bring loaded to, or above, its yield stress. Yielding is only one of many possible failure mechanisms, and our analysis in this section will not offer predictions regarding other kinds of nature. When a straight rod is placed in tension as in figure 5.2, the possibility of it yielding is assessed by comparing the stress, , to the materials yield strength. Failure due to ductile yielding is predicted if Sy . Engineers define the tensile factor of safety as Ntension=sy . If n 1, this viewpoint predicts that the component will not yield, and if n0) when the object moves higher(h0). Conversely, the gravitational potential energy decreases (U0) when the object is lowered (h0) because the force acts in the same direction as the pistons motion. Conversely, the work is negative if the force opposes motion, as in Figure 6.2(b) with d 0.Power, the last of the quantities introduced in this section, is defined as the rate at which work is performed. When a force performs work during the interval of timet .the average power is defined：Pavg=W /t （6.5） As work is performed more rapidly, t becomes smaller, and the average power likewise increases. Engineers conventionally express power in the units of watt (1W=1J/s) in the SI, and either ftlb/s or horsepower（hp）in the USCS. Table 6.2 lists conversion factors between those choices of units. Reading off the third row, for instance, we see that horsepower is converted to ftlb/s according to 1 hp=550 ftlb/s.Example 6.1 Beginning with the definitions of ft.lb/s and W,determine the conversion factor between them in table Solution:1ft.lb/s=(1ft.lb/s)(0.3048m/ft)(4.448N/lb)=1.356N.m/s=1.356W1ft.lb/s=(1ft.lb/s)(0.3048m/ft)(4.448N/lb)=1.356N.m/s=1.356WExample 6.2 Calculate the elastic potential energy stored in the two straight sections of the U-bolt examined in Example 5.1 and 5.2.Solution:With the dimensions and material properties used in those examples, the spring constant for one straight section is found from equation 5.5:=(210109Pa)(7.8510-5m2)/0.3m=5.5107N/mIn example 5.2,each straight section of the U-bolt was found to stretch by 72.6 m.by using equation 6.2, the elastic potential energy becomes :U=(5.5107N/m)(7.2610-5m ) 2/2 =0.145N.m=0.145JBecause the U-bolt has two identical straight section , the stored energy is 2(0.145J)=0.29J,a relatively modest amount of energy.Example 6.3 In September of 2001, terrorists used commercial jetliners to destroy the twin towers of the World Trade Center in New York City. Calculate the kinetic energy of a Boeing 767 that is loaded to its maximum weight of 350,000 lb and travels at 400 mph.Solution: To apple equation 6.3,we must fist calculate the aircrafts mass:m=(3.5105 lb)/(32.2ft/s2)=1.087104 slugsAnd convert speed into the consistent units of ft/s:V=(400mph)(5280ft/mi)(2.77810-4 ft/s) 2=578 ft/sThe kinetic energy becomes :U=(1.087104 slugs) (578 ft/s) 2/2=2.54109J=2.54GJin the SI . To place that quantity of energy into perspective ,we recall from Problem 18 in Chapter 2 that the unit “kiloton” is sometimes used to describe the energy produced by large explosions .Being approximately equivalent to the explosive energy of 1000 tons of high explosive , the kiloton is defined as . With that conversion factor ,we see that the aircrafts kinetic energy is comparable to 6*10-4kiloton ,or some 1200 lb of high explosives.Example 6.4 Make an order- of- magnitude calculation to estimate the capacity ,in the unites of horsepower ,of an electric motor that will power an elevator in a four-story office building. The elevator car weights 500lb , and it will carry up to an additional 2500 lb of passengers and freight.Solution We apply Equation 6.5 to estimate the motors power rating. We estimate that it takes the elevator 20 seconds to travel from ground level to the top floor and that the total elevation change is 50 ft. With those assumptions, the average power is:Pavg=W/t =(500lb+2500lb)(50ft)/20s=7500ft.lb/s Converting to the unit of horsepower, P= (7500 ft.lb/s)(1.81810-3 hp/ (ft.lb/s)=13.6 hpAs an engineer later accounts for friction, inefficiency of the motor, the possibility of the elevator being overloaded, and a factor of safety, the calculation would become more accurate. From a preliminary standpoint, however, a motor rated for a few tens of horsepower, rather than one with several horsepower or several hundred, would be sufficient.6.3 Heat as energy in transit In the previous section , we considered several types of energy that can be stored within a mechanical system .Energy can also be converted from one from to another for instance , when the potential energy of a falling object is transferred to its kinetic energy .We view heat as energy that is in transit from one location to another because of a temperature difference .In this section ,we explore several engineering concepts that are related to heat ,its release when a fuel burns ,and its transfer through the processes known as conduction , convection , and radiation. When a fuel such as oil (liquid),coal(solid),or propane(gas)is burned ,the chemical reactions that occur release heat and by-products including water vapor ,carbon monoxide , and particulates .The chemical energy that is liberated as heat during combustion is measured by a quantity called the heat of combustion ,H. As listed in Table 6.3 ,the numerical values for the heat of combustion describe the ability of fuels to produce heat ,and H is specified in the units MJ/kg in the SI and Btu/slug in the USCS .Conversion factors between conventional units in the USCS and SI for heat of combustion are listed in Table 6.4 .The unit Btu shown in Tables 6.1 and 6.3 defined as the quantity of heat that must be supplied in order to raise the temperature of 1 lb of water by1。F. In the modern definition , the Btu is equivalent to 778.2ft.lb or 1055J . The Btu is a commonly used unit for heat energy calculations in mechanical engineering .As a general rule ,the heatQ that is released when a mass m of fuel is burned is given by：Q= m H (6.6)Referring to Table 6.3, when 1 kilogram of gasoline is burned ,some 45 MJ of heat are released .Equivalently in the USCS ,we say that 6.25105 Btu are released when 1 slug of gasoline is consumed . If we could somehow build an automobile engine that perfectly converts that quantity of heat into kinetic energy ,a 1000-kilogram vehicle could be accelerated to a speed of nearly 300m/s ,roughly the speed of sound at sea level .Table 6.3 heat of combustion values for different fuelsheat of combustion ,HtypeSubstance MJ/kgBtu/sluggasNatural gas 253.5105liquidmethanol233.2105ethanol304.2105gasoline456.2105Fuel oil425.8105Solid coal283.9105wood202.8105Table 6.4 conversion factors between units for heat of combustion in the USCS and SIMJ/kgBtu/slug11.3831047.22910-51Example 6.5 A gasoline-powered internal combustion engine generates an average power output of 50kW (about 67 hp in the USCS ). Neglecting the inefficiency of the engine ,calculate the volume of fuel that is consumed in 1 hour . Express your result in both the SI and USCS .Solution：By discounting inefficiency , we recognized that our calculation will underestimate the actual rate of fuel consumption。In 1 hour ,the engine produces an energy output of 50Kw1 hour=(50kJ/s)(3600s)=1.8105 kJ=180MJReferring to Table 6.3 the heat of combustion for gasoline is 45 MJ/kg .Applying Equation 6.6 ,the mass of fuel burned ism=180 MJ /45 MJ/kg=4 kgIn Table 4.3 ,the density of gasoline is listed as 680kg/m3,and so the engine burns :V=4 kg/680kg/m3=0.0059 m3=5.9Lwhere the conversion 1 m3 =1000L from Table 2.4 was used .in the USCSand with the conversion factor 1L=0.2642gal from table 2.7，1.6 gal of fuel is consumed in 1 hour.Example 6.6 The average single-family household in the united states consumes 98 million btu of energy each year .In the units of lb ,what quantity of fuel oil must be burned to produce that amount of energy ? Solution:The mass of fuel oil is found by applying Equation 6.6: m=98 106 Btu/5.8105 Btu /slugFrom which m=1
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