关 键 词：

机械类毕业-含CAD图纸 风机 机构 设计 机械类 毕业 cad 图纸

资源描述：

竖轴风机变角机构设计【机械类毕业-含CAD图纸】,机械类毕业-含CAD图纸,风机,机构,设计,机械类,毕业,cad,图纸

内容简介：

设计（论文）题目：竖轴风机变角机构设计 任务书填写要求1毕业设计（论文）任务书由指导教师根据各课题的具体情况填写，经学生所在专业的负责人审查、系（院）领导签字后生效。此任务书应在毕业设计（论文）开始前一周内填好并发给学生。2任务书内容必须用黑墨水笔工整书写，不得涂改或潦草书写；或者按教务处统一设计的电子文档标准格式（可从教务处网页上下载）打印，要求正文小4号宋体，1.5倍行距，禁止打印在其它纸上剪贴。3任务书内填写的内容，必须和学生毕业设计（论文）完成的情况相一致，若有变更，应当经过所在专业及系（院）主管领导审批后方可重新填写。4任务书内有关“学院”、“专业”等名称的填写，应写中文全称，不能写数字代码。学生的“学号”要写全号，不能只写最后2位或1位数字。 5任务书内“主要参考文献”的填写，应按照金陵科技学院本科毕业设计（论文）撰写规范的要求书写。6有关年月日等日期的填写，应当按照国标GB/T 740894数据元和交换格式、信息交换、日期和时间表示法规定的要求，一律用阿拉伯数字书写。如“2002年4月2日”或“2002-04-02”。毕 业 设 计（论 文）任 务 书1本毕业设计（论文）课题应达到的目的： 本课题属于教师自主命题，来源于工程实践。目的：1、通过本课题的设计研究，考察学生四年来在校所学的专业知识水平及运用专业知识解决设计项目的创新能力；2、通过本课题的研究使学生系统的熟悉机械设计分析及掌握相关的设计手法。 3、通过本课题使学生熟练掌握制图方法、规范设计图纸画法以及提高使用设计软件解决应用问题的能力。 2本毕业设计（论文）课题任务的内容和要求（包括原始数据、技术要求、工作要求等）： 内容：本课题为垂直轴风机叶片变偏角机构设计，本设计力求在保证风机旋的基础上，同时满足叶片自身的转动，并控制叶片转动的角度，以免造成叶片损伤，从动力装置、保护措施、外形等方面打造变偏角机构设计，创造出新型垂直轴风机。本课题技术要求：1、创造性：设计应为原创。设计作品应具有较强的创新内涵，注重设计思考与设计表达的统一。2、功能性：注重实用功能，在外形美观的同时发挥实用功效。3、合理性：垂直走风机叶片的设计要考虑叶片受力情况，并从这方面入手合理选择材料及设计手法。 本课题研究的工作要求：1、前期准备：查找国内外垂直轴风机实例，研究叶片部分结构。2、查阅文献资料：风力发电、风机翼型研究、材料力学等相关书籍资料和网站。完成设计方案，设计图纸。 毕 业 设 计（论 文）任 务 书3对本毕业设计（论文）课题成果的要求包括图表、实物等硬件要求： 1、设计图纸： 装配图，零号图2、设计说明书。4主要参考文献： 1王承煦，张源.风力发电M.北京：中国电力出版社，2002.2廖明夫，R.Gasch，J.Twele.风力发电技术M.西安:西北工业大学出版社，2009.3李俊峰等.中国风电发展报告M.北京：中国环境科学出版社，2012.4郭太英，黎发贵.从国外风电发展探讨我国风电发展思路J.水电勘测设计，2006.5黄继雄.风力机专用新翼型及其气动特性研究:D.汕头大学，20016唐进.提高风力机翼型气动性能的研究:D.清华大学，20047沈观林，胡更开.复合材料力学M.北京：清华大学出版社，2006.8赵美英，陶梅贞.复合材料结构力学与结构设计M.西安：西北工业大学出版社，20079美国通用电气公司.多段式风力涡轮机叶片和用于组装该叶片的方法P.中国专利:200810188672，2009-06-24.10张石强.风力机专用翼型及叶片关键设计理论研究D.重庆：重庆大学，2010.11刘雄，陈严，叶枝全.风力机桨叶总体优化设计的复合形法J.太阳能学报2001,22(2):157-161.12Tony Burtyon著，武鑫译.风能技术M.北京：科学技术出版社，200713包耳，邵晓荣，刘德庸.风力机叶片设计的新方法J.机械设计，2005,22(2)：24-2614赵丹平.风力发电机组叶片模型气动载荷研究D.内蒙古农业大学200915黄华.风力机风轮叶片设计计算方法研究D.西华大学200816J.Selwin Rajadurai,T.Christopher,G.Thanigaiyarasu3B-Nageswara Rao.Finiteelement analysis with an improved failure criterion for composite windturbine blades J.Forschung im Ingenieurwesen，2008，72(4): 193-207.17M.Grujicic，G.Arakere，E.Subramanian，V.Sellappan，A.Vallejo，M.Ozen.Structural-Response Analysis,Fatigue-Life Prediction,and Material Selection for 1MWHorizontal-Axis wind-Turbine BladesJ.Materials Engineering and Performance, 2010,19(6):790-801.18Jayantha A.Epaarachchi,Philip D. Clausen.The development of a fatigueloading spectrum for small wind turbine bladesJ Journal of wind engineeringand industrial aerodynamics，2006，94(4):207-223.19J.C.Marin,A.Barroso,F.Paris,J.Carias.Study of fatigue damage in wind tubineblades J . Engineering Failure Analysis,2009,16(2):656-666.20Changduk Kong，Taekhyun Kim,Dongju Han，Yoshihiko Sugiyama.Investigation of fatigue life for a medium scale composite wind turbine bladeJ Journal of Fatigue,2006,28(10):1382-1388.119毕 业 设 计（论 文）任 务 书5本毕业设计（论文）课题工作进度计划：15.11.20-15.12.2015.12.20-16.01.1516.01.15-16.03.1816.03.18-16.04.0816.04.08-16.04.3016.05.01-16.05.1016.05.10-16.05.15学生明确选题学生完成开题报告学生完成设计草图阶段,明确设计方案学生完善设计正稿, 撰写毕业设计论文初稿学生毕业设计完成阶段,提交毕业论文正稿,完成期中检查学生提交毕业设计论文,布置毕业设计展布展、毕业答辩准备所在专业审查意见：通过负责人： 2015 年 12 月23 日 毕 业 设 计（论文） 开 题 报 告 1结合毕业设计（论文）课题情况，根据所查阅的文献资料，每人撰写不少于1000字左右的文献综述： 新兴市场的风电发展迅速，在国家政策支持和能源供应紧张的背景下，中国的风电特别是风电设备制造业也迅速崛起，已成为全球风电最为活跃的场所。2006年全球风电资金中9%投向了中国，总额高达16.2亿欧元。2007年，中国风电装机容量已排名世界第五。截止到2012年，中国风电装机容量达到42287MW，跃居世界第一。 1.风力发电机的分类 1.1根据轴的状态分类：尽管风力发电机的种类多种多样，但总体来说可以归纳为两类：1.水平轴风力发电机；2.垂直轴风力发电机；3.s型风车。这其中性能最为优越的当属垂直轴风力发电机，首先，它无须在风向改变时进行对风，这一点比水平轴风机简便；第二，输出功率高，这又是s型风车所欠缺的。总结来说，垂直轴风力发电机不仅结构设计简化，且减少了风轮对风时的陀螺力。 1.2垂直轴风力发电机分类：垂直轴风力发电机还可以进行再分类，其中包括：1.达里厄式风轮。是由法国人G.J.M达里厄于19世纪30年代发明的。在20世纪70年代，加拿大国家科学研究院对此进行了大量的研究，是水平轴风力发电机的主要竞争者。达里厄式风轮是一种升力装置，弯曲叶片的剖型是翼型，它的启动力矩低，但尖速比可以很高，对于给定的风轮重量和成本，有较高的输出。这些风轮可以设计成单叶片，双叶片，三叶片或者多叶片。其他形式的垂直轴风力发电机有马格努斯效应风轮，他由自旋的圆柱体构成，当它在气流中工作时，产生的移动力是由马格努斯效应引起的，其大小与风速成正比。有的垂直轴风轮使用管道或者漩涡发生器塔，通过套管或者扩压器使水平气流变成垂直气流，以增加速度，偶尔还利用太阳能或者燃烧某种燃料，使水平气流变成垂直气流。垂直轴风力发电机发展到目前为止已经经历了很大的变化，随着科技的进步，目前已出现了一种新型的垂直轴风力发电机。该种发电机采用空气洞力学原理，针对垂直轴旋转的风洞模拟，叶片选用了飞机翼型形状，在风轮旋转时，它不会受到因变形而改变效率等；它用垂直直线4-5个叶片组成，由4角形或5角形形状的轮毂固定，连接叶片的连杆组成的风轮，由风轮带动稀土永磁发电机发电送往控制器进行控制，输配负载所用的电能。该技术原理根据空气片条理论，实际计算可选取垂直轴风力发电机旋转轴的切面进行计算模型，按叶片的实际尺寸，每个叶片的旋转轴心距为N米；用CFD技术进行模拟气动系数计算，计算原理采用离散数字方法求解翼形断面的气动力，用网格方法对雷诺数流动涡量分布比较形成高雷诺数下对Navier-Stokes方程进行数字模拟计算的原理结果。采用稀土永磁材料发电的原理，配套空气洞力学原理的风轮，采用直驱式结构进行选装发电。 1.3根据驱动方式分类：同时，还可分为阻力型和升力型。阻力型垂直轴风力发电机主要是利用空气流过叶片产生的阻力作为驱动力的，而升力型则是利用空气流过叶片产生的升力作为驱动力的。由于叶片在旋转过程中，随着转速的增加阻力急速减小。而升力反而会增大，所以升力型的垂直轴风力发电机的效率要比阻力型搞得多。 2.垂直轴风力发电机的未来发展风力发电和太阳能发电一样，最初的目的是为了解决应急电源和边远地区供电而开发出来的产品，因而在最初的发展并不是很快。到了上个世纪二，三十年代，全球经济危机带来的能源紧张，让世界各国的专家想到了以风力发电作为补充能源的可能性，二次世界大战后，各国纷纷进行研究，由于当时的技术水平有限，启动风速要求较高，发电噪音也很大，所以只能将风力发电机放在人迹罕至的地方或风力较大的地方。设备也往往是往大型风力发电机发展，专门建设大型风力发电机厂，由于水平轴风力发电机的特性，小型风力发电在相当长的时间里未得到较好的发展。各个发达国家均致力于发展新兴能源产业，而太阳能一直是新兴能源商业化的首选，因为太阳能的设置地点较灵活，不会产生噪音，可以和建筑进行一体化设计，在建筑周围设置小型风力发电机而又不影响人的生活质量，这成了欧美一些国家研究的焦点。 3.垂直轴风力发电机发展现状 2002年，中国率先开始了新型垂直轴风力发电机的研究，由部队通讯部牵头，上海某公司为研发主体，西安军电，西安交大，同济大学，复旦大学等搞笑的多位专家配合，在短短一年时间里就产生了首台新型垂直轴风力发电机。并在不到5年的时间里将功率扩展至200W100KW，处于世界领先地位。按照我国“十二五”规划目标，预计到2015年风力发电机容量将达到1*108KW，年发电量1900*108KW.h。GWEC和Greenpeace预测，今后20年风力发电将成为世界主力电源，2030年装机容量有可能达到23*108KW，可供应世界电力需求的22%。 毕 业 设 计（论文） 开 题 报 告 2本课题要研究或解决的问题和拟采用的研究手段（途径）： 1.明确设计要求 参考其他同类型设计，从形状，原理尺寸精度等方面考虑设计的可行性与经济性。 2.运用CAD及Creo软件完成设计 零件图尺寸合理性，装配图零件与零件间的连接等方面需特别注意认真绘制。 3.结构设计 查阅有关空气洞力学相关书籍，综合考虑零件消耗和功率问题。 4.总体尺寸的确定 参考其他设计，查阅相关书籍进行计算。 毕 业 设 计（论文） 开 题 报 告 指导教师意见：1对“文献综述”的评语：该生通过大量搜集和查阅文献资料，对本课题相关的国内外前人工作较好地进行了综合分析和归纳整理，并针对某一学者具体的研究工作进行了比较专门的、全面的、深入的和系统的描述与评价，语言简洁，层次清楚。达到了学校“文献综述要求”。 2对本课题的深度、广度及工作量的意见和对设计（论文）结果的预测：预期成果可以完成3.是否同意开题： 同意 不同意 指导教师： 2016 年 03 月 19 日所在专业审查意见：同意 负责人： 2016 年 04 月 05 日译文题目： 材料力学的分析与设计 说明：要求学生结合毕业设计（论文）课题参阅一篇以上的外文资料，并翻译至少一万印刷符（或译出3千汉字）以上的译文。译文原则上要求打印（如手写，一律用400字方格稿纸书写），连同学校提供的统一封面及英文原文装订，于毕业设计（论文）工作开始后2周内完成，作为成绩考核的一部分。OBJECTIVESThe main objective of a basic mechanics course should be to develop in the engineering student the ability to analyze a given problem in a simple and logical manner and to apply to its solution a few fundamental and well-understood principles. This text is designed for the first course in mechanics of materialsor strength of materials offered to engineering students in the sophomore or junior year. The authors hope that it will help instructors achieve this goal in that particular course in the same way that their other texts may have helped them in statics and dynamics. GENERAL APPROACH In this text the study of the mechanics of materials is based on the understanding of a few basic concepts and on the use of simplified models. This approach makes it possible to develop all the necessary formulas in a rational and logical manner, and to clearly indicate theconditions under which they can be safely applied to the analysis and Designation nonfactual restructurings and machine components.Design Concepts Are Discussed Throughout the Text When-ever Appropriate. A discussion of the application of the factor of safety to design can be found in Chap. 1, where the concepts of both allowable stress design and load and resistance factor design are presented.A Careful Balance Between SI and U.S. Customary Units Is Consistently Maintained. Because it is essential that students be able to handle effectively both SI metric units and U.S. customary units, half the examples, sample problems, and problems to be assigned have been stated in SI units and half in U.S. customary units. Since a large number of problems are available, instructors can assign problems using each system of units in whatever proportion they find most desirable for their class.IntroductionThe main objective of the study of the mechanics of materials is to provide the future engineer with the means of analyzing and designing various machines and load-bearing structures.Both the analysis and the design of a given structure involve the determination of stresses and deformations. This first chapter is devoted to the concept of stress.Section 1.2 is devoted to a short review of the basic methods of statics and to their application to the determination of the forces in the members of a simple structure consisting of pin-connected members.Section 1.3 will introduce you to the concept of stress in a member of a structure, and you will be shown how that stress can be determined from the force in the member. After a short discussion of engineering analysis and design (Sec. 1.4), you will consider successively the normal stresses in a member under axial loading (Sec. 1.5), the shearing stresses caused by the application of equal and opposite transverse forces (Sec. 1.6), and the bearing stresses created by bolts and pins in the members they connect (Sec. 1.7). These various concepts will be applied in Sec. 1.8 to the determination of the stresses in the members1.2A SHORT REVIEW OF THE METHODS OF STATICSIn this section you will review the basic methods of statics while determining the forces in the members of a simple structure.Consider the structure shown in Fig. 1.1, which was designed to support a 30-kN load. It consists of a boom AB with a 30 3 50-mm rectangular cross section and of a rod BC with a 20-mm-diameter circular cross section. The boom and the rod are connected by a pin at B and are supported by pins and brackets at A and C, respectively. Our first step should be to draw a free-body diagram of the structure by detaching it from its supports at A and C, and showing the reactions that these supports exert on the structure (Fig. 1.2). Note that the sketch of the structure has been simplified by omitting all unnecessary details. Many of you may have recognized at this point that AB and BC are two-force members. For those of you who have not, we will pursue our analysis, ignoring that fact and assuming that the directions of the reactions at A and C are unknown. Each of these reactions, therefore, will be represented by two components, Ax and Ay at A, and Cx and Cy at C. We write the following three equilibrium equations:We have found two of the four unknowns, but cannot determine the other two from these equations, and no additional independent equation can be obtained from the free-body diagram of the structure. We must now dismember the structure. Considering the free-body diagram of the boom AB (Fig. 1.3), we write the following equilibrium equation:Substituting for Ay from (1.4) into (1.3), we obtain Cy=+30 kN. Expressing the results obtained for the reactions at A and C in vector form, we haveWe note that the reaction at A is directed along the axis of the boom AB and causes compression in that member. Observing that the components Cx and Cy of the reaction at C are, respectively, proportional to the horizontal and vertical components of the distance from B to C, we conclude that the reaction at C is equal to 50 kN, is directed along the axis of the rod BC, and causes tension in that memberThese results could have been anticipated by recognizing that AB and BC are two-force members, i.e., members that are subjected to forces at only two points, these points being A and B for member AB, and B and C for member BC. Indeed, for a two-force member the lines of action of the resultants of the forces acting at each of the two points are equal and opposite and pass through both points. Using this property, we could have obtained a simpler solution by considering the free-body diagram of pin B. The forces on pin B are the forces FAB and FBC exerted, respectively, by members AB and BC, and the 30-kN load (Fig. 1.4a). We can express that pin B is in equilibrium by drawing the corresponding force triangle (Fig. 1.4b).Since the force FBC is directed along member BC, its slope is the same as that of BC, namely, 3/4.We can, therefore, write the proportion:from which we obtain:The forces F9AB and F9BC exerted by pin B, respectively, on boom AB and rod BC are equal and opposite to FAB and FBC (Fig. 1.5).Knowing the forces at the ends of each of the members, we can now determine the internal forces in these members. Passing a section at some arbitrary point D of rod BC, we obtain two portions BD and CD (Fig. 1.6). Since 50-kN forces must be applied at D to both portions of the rod to keep them in equilibrium, we conclude that an internal force of 50 kN is produced in rod BC when a 30-kN load is applied at B. We further check from the directions of the forces Fbc and Fbc in Fig. 1.6 that the rod is in tension. A similar procedure would enable us to determine that the internal force in boom AB is 40 kN and that the boom is in compression.1.3 STRESSES IN THE MEMBERS OF A STRUCTUREWhile the results obtained in the preceding section represent a first and necessary step in the analysis of the given structure, they do not tell us whether the given load can be safely supported. Whether rod BC, for example, will break or not under this loading depends not only upon the value found for the internal force FBC, but also upon the cross-sectional area of the rod and the material of which the rod is made. Indeed, the internal force FBC actually represents the resultant of elementary forces distributed over the entire area A of the cross section (Fig. 1.7) and the average intensity of these distributed forces is equal to the force per unit area, FBCyA, in the section.Whether or not the rod will break under the given loading clearly depends upon the ability of the material to withstand the corresponding value FBCyA of the intensity of the distributed internal forces. It thus depends upon the force FBC, the cross-sectional area A, and the material of the rod.The force per unit area, or intensity of the forces distributed over a given section, is called the stress on that section and is denoted by the Greek letter s (sigma). The stress in a member of cross-sectional area A subjected to an axial load P (Fig. 1.8) is therefore obtained by dividing the magnitude P of the load by the area A:A positive sign will be used to indicate a tensile stress (member in tension) and a negative sign to indicate a compressive stress (member in compression).Since SI metric units are used in this discussion, with P expressed in newtons (N) and A in square meters (m2), the stress swill be expressed in N/m2. This unit is called a pascal (Pa). However, one finds that the pascal is an exceedingly small quantity and that, in practice, multiples of this unit must be used, namely, the kilopascal (kPa), the megapascal (MPa), and the gigapascal (GPa).We haveWhen U.S. customary units are used, the force P is usually expressed in pounds (lb) or kilo-pounds (kip), and the cross-sectional area A in square inches (in2). The stress s will then be expressed in pounds per square inch (psi) or kilo-pounds per square inch (ksi).1.4ANALYSIS AND DESIGNConsidering again the structure of Fig. 1.1, let us assume that rod BC is made of a steel with a maximum allowable stress all =165 MPa. Can rod BC safely support the load to which it will be subjected? The magnitude of the force FBC in the rod was found earlier to be 50 kN. Recalling that the diameter of the rod is 20 mm, we use Eq. (1.5) to determine the stress created in the rod by the given loading. We haveSince the value obtained for s is smaller than the value sall of the allowable stress in the steel used, we conclude that rod BC can safely support the load to which it will be subjected. To be complete, our analysis of the given structure should also include the determination of the compressive stress in boom AB, as well as an investigation of the stresses produced in the pins and their bearings. This will be discussed later in this chapter. We should also determine whether the deformations produced by the given loading are acceptable. The study of deformations under axial loads will be the subject of Chap. 2.An additional consideration required for members in compression involves the stability of the member, i.e., its ability to support a given load without experiencing a sudden change in configuration. This will be discussed in Chap. 10.The engineers role is not limited to the analysis of existing structures and machines subjected to given loading conditions. Of even greater importance to the engineer is the design of new structures and machines, that is, the selection of appropriate components to perform a given task. As an example of design, let us return to the structure of Fig. 1.1, and assume that aluminum with an allowable stress all= 5 100 MPa is to be used. Since the force in rod BC will still be P 5 FBC 5 50 kN under the given loading, we must have,from Eq. (1.5),We conclude that an aluminum rod 26 mm or more in diameter will be adequate.AXIAL LOADING; NORMAL STRESSAs we have already indicated, rod BC of the example considered in the preceding section is a two-force member and, therefore, the forces Fbc and Fbc acting on its ends B and C (Fig. 1.5) are directed along the axis of the rod. We say that the rod is under axial loading. An actual example of structural members under axial loading is provided by the members of the bridge truss shown in Photo 1.1.Returning to rod BC of Fig. 1.5, we recall that the section we passed through the rod to determine the internal force in the rod and the corresponding stress was perpendicular to the axis of the rod; the internal force was therefore normal to the plane of the sec-tion (Fig. 1.7) and the corresponding stress is described as a normal stress. Thus, formula (1.5) gives us the normal stress in a member under axial loading:We should also note that, in formula (1.5), s is obtained by dividing the magnitude P of the resultant of the internal forces distributed over the cross section by the area A of the cross section; it represents, therefore, the average value of the stress over the cross section, rather than the stress at a specific point of the cross section.To define the stress at a given point Q of the cross section, we should consider a small area DA (Fig. 1.9). Dividing the magnitude of DF by DA, we obtain the average value of the stress over DA. Letting DA approach zero, we obtain the stress at point Q:In general, the value obtained for the stress s at a given point Q of the section is different from the value of the average stress given by formula (1.5), and s is found to vary across the section. In a slender rod subjected to equal and opposite concentrated loads P and P9 (Fig. 1.10a), this variation is small in a section away from the points of application of the concentrated loads (Fig. 1.10c), but it is quite noticeable in the neighborhood of these points (Fig. 1.10b and d).But the conditions of equilibrium of each of the portions of rod shown in Fig. 1.10 require that this magnitude be equal to the mag-nitude P of the concentrated loads. We have, therefore,which means that the volume under each of the stress surfaces inFig. 1.10 must be equal to the magnitude P of the loads. This, however, is the only information that we can derive from our knowledge of statics, regarding the distribution of normal stresses in the various sections of the rod. The actual distribution of stresses in any given section is statically indeterminate. To learn more about this distribution, it is necessary to consider the deformations resulting from the particular mode of application of the loads at the ends of the rod.This will be discussed further in Chap. 2.In practice, it will be assumed that the distribution of normalstresses in an axially loaded member is uniform, except in the immediate vicinity of the points of application of the loads. The value of the stress is then equal to and can be obtained from formula(1.5). However, we should realize that, when we assume a uniform distribution of stresses in the section, i.e., when we assume that the internal forces are uniformly distributed across the section, it follows from elementary statics that the resultant P of the internal forces must be applied at the centroid C of the section (Fig. 1.11). This means that a uniform distribution of stress is possible only if the line of action of the concentrated loads P and P9 passes through the centroid of the section considered (Fig. 1.12). This type of loading is called centric loading and will be assumed to take place in all straight two-force members found in trusses and pin-connected structures, such as the one considered in Fig. 1.1. However, if a two-force member is loaded axially, but eccentrically as shown in Fig. 1.13a, we find from the conditions of equilibrium of the portion of member shown in Fig. 1.13b that the internal forces in a given section must be equivalent to a force P applied at the centroid of the section and a couple M of moment M 5 Pd. The distribution of forcesand, thus, the corresponding distribution of stressescannot be uniform. Nor can the distribution of stresses be symmetric as shown in Fig. 1.10. This point will be discussed in detail in Chap. 4. 材 料 力 学一个基本的力学课程的主要目的应该是提供给工科学生发展到分析给定问题一个简单而合乎逻辑的方式，并应用到其解决一些基本的和易于理解的原则的能力。本文是专业材料，是为大二或大三工科学生提供的材料 - 强度力学第一期课程。作者希望这将以同样的方式帮助教师实现这个目标，其他文章可以在静态和动态两方面也能帮助他们。在这段文字材料中，力学的研究是基于几个基本概念的理解和使用简化的模型。这种方法使得有可能开发出理性和合乎逻辑的方式的所有必要公式，在清楚地指示下，他们可以安全地施加到分析和指定非事实重组和机器部件的条件。设计概念在整个文本中只要是合理的位置都进行了讨论。安全设计的要素应用的讨论可以在11.2节找到。无论在哪里许用应力设计载荷和阻力系数设计的概念介绍都可以找到。SI和美国习惯单位之间的谨慎平衡始终保持。 因为它是必不可少的学生能够有效地处理这两种SI公制单位和美国惯用单位，有一半的例子说明，采样的问题，和要分配在SI单位和一半的美国惯用单位已陈述的问题。由于大量的问题是可用的，教师可以以仍宜比例分配使用的，他们发现最希望为他们发现的类单元，每个单元的系统问题。材料力学的研究的主要目标是提供一种具有分析和设计各种机器和承重结构的装置能力的未来工程师。既分析并设计一个给定结构的涉及应力和变形的测定。这第一章介绍应力的概念。1.2节所专门介绍的是静力学和他们在由简单结构的成员力量的中心的基本方法连接方面的应用。1.3节将向您介绍应力的一个概念结构的成员，你将看到压力如何可以从该成员的力量来决定。在工程分析和设计（1.4节）的一个简短的讨论后，你会陆续考虑成员在轴向载荷（1.5节）的正应力，造成相等，方向相反的横向力的应用程序（1.6节）的剪切应力，并通过螺栓和销的成员的连接所创建的承载压力（1.7节）。这些不同的概念将在1.8节中被应用，并对应力的成员进行判断。有关静力学的简短评述在本节中，你将会学习到一种确定简单结构各部分受力的基本静力学方法。考虑图1.1所示的结构，它被设计为可以支持30千牛的负荷的载体。它由一个用30* 50毫米矩形横截面的悬臂AB和直径20毫米的圆形横截面的杆组成。起重臂和杆通过在B中的销连接，并且通过销和支架在A和C，分别支承。我们的第一个步骤应该是从其在A和C支持之中分离，再将其表明这些支持在结构（图1.2）发挥反应绘制结构的自由体图。需要注意的是该结构的草图已经通过省略所有不必要的细节简化。很多人可能在这一点上已经认识到，AB和BC是两个力的成员。对于那些你们没有，我们将继续我们的分析，忽略了这一事实，并假设在A和C反应的方向是未知的。因此，这些反应中的每一个都将由两部分组成，Ax和Ay在A和Cx和Cy在C.我们写以下三个平衡方程来表示我们已经发现了两个的四个未知数，但并不能从这些方程中确定其它两个，并且也不能从该结构的自由体图获得额外的独立方程。我们现在必须拆分结构。考虑到杆AB（图1.3），我们编写以下平衡方程的自由体图 将Ay从（1.4）代入到（1.3），我们可以得到Cy=+30千牛。将它表示为在A和C中的矢量形式的反应所获得的结果，我们有我们注意到，在A中的反应是沿着吊杆AB的轴线定向并且使得压缩该构件。观察该元件Cx和中C反应Cy分别是，正比于从B到C的距离的水平和垂直分量，我们得出结论，在C反应等于50千牛，沿轴线定向杆BC，并导致该成员的张力这些结果可能已经通过识别AB和BC两种力件，即，是在只有两点承受力的成员，这些点是A和B构件AB和B和C件BC的预期。的确，对于两力构件在每个两分的作用的力的合力的作用线是相等且相反的并穿过两个点。利用这个特性，我们可以考虑销B的自由体图上针B上的力量是FAB和FC施加的力，分别由成员AB和BC，以及30千牛负荷（获得一个简单的解决方案图1.4A）。我们可以表达该引脚B是平衡通过绘制相应的力三角形（图1.4B）。既然