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畢 業(yè) 設(shè) 計(jì)（論 文）外 文 參 考 資 料 及 譯 文
譯文題目： 材料力學(xué)的分析與設(shè)計(jì)
學(xué)生姓名： 學(xué) 號(hào)：
專(zhuān) 業(yè)：
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20xx年 2月 27日
說(shuō)明：
要求學(xué)生結(jié)合畢業(yè)設(shè)計(jì)（論文）課題參閱一篇以上的外文資料，并翻譯至少一萬(wàn)印刷符（或譯出3千漢字）以上的譯文。譯文原則上要求打印（如手寫(xiě)，一律用400字方格稿紙書(shū)寫(xiě)），連同學(xué)校提供的統(tǒng)一封面及英文原文裝訂，于畢業(yè)設(shè)計(jì)（論文）工作開(kāi)始后2周內(nèi)完成，作為成績(jī)考核的一部分。
OBJECTIVES
The 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 materials—or 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 the
conditions 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.
Introduction
The 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 members
1.2A SHORT REVIEW OF THE METHODS OF STATICS
In 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 have
We 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 member
These 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 F’bc 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 STRUCTURE
While 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 have
When 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).
1.4 ANALYSIS AND DESIGN
Considering 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 have
Since 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 engineer’s 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 STRESS
As 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 F’bc 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 in
Fig. 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 normal
stresses 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 forces—and, thus, the corresponding distribution of stresses—cannot 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 學(xué)
一個(gè)基本的力學(xué)課程的主要目的應(yīng)該是提供給工科學(xué)生發(fā)展到分析給定問(wèn)題一個(gè)簡(jiǎn)單而合乎邏輯的方式，并應(yīng)用到其解決一些基本的和易于理解的原則的能力。本文是專(zhuān)業(yè)材料，是為大二或大三工科學(xué)生提供的材料 - 強(qiáng)度力學(xué)第一期課程。作者希望這將以同樣的方式幫助教師實(shí)現(xiàn)這個(gè)目標(biāo)，其他文章可以在靜態(tài)和動(dòng)態(tài)兩方面也能幫助他們。
在這段文字材料中，力學(xué)的研究是基于幾個(gè)基本概念的理解和使用簡(jiǎn)化的模型。這種方法使得有可能開(kāi)發(fā)出理性和合乎邏輯的方式的所有必要公式，在清楚地指示下，他們可以安全地施加到分析和指定非事實(shí)重組和機(jī)器部件的條件。
設(shè)計(jì)概念在整個(gè)文本中只要是合理的位置都進(jìn)行了討論。安全設(shè)計(jì)的要素應(yīng)用的討論可以在11.2節(jié)找到。無(wú)論在哪里許用應(yīng)力設(shè)計(jì)載荷和阻力系數(shù)設(shè)計(jì)的概念介紹都可以找到。
SI和美國(guó)習(xí)慣單位之間的謹(jǐn)慎平衡始終保持。
因?yàn)樗潜夭豢缮俚膶W(xué)生能夠有效地處理這兩種SI公制單位和美國(guó)慣用單位，有一半的例子說(shuō)明，采樣的問(wèn)題，和要分配在SI單位和一半的美國(guó)慣用單位已陳述的問(wèn)題。由于大量的問(wèn)題是可用的，教師可以以仍宜比例分配使用的，他們發(fā)現(xiàn)最希望為他們發(fā)現(xiàn)的類(lèi)單元，每個(gè)單元的系統(tǒng)問(wèn)題。
材料力學(xué)的研究的主要目標(biāo)是提供一種具有分析和設(shè)計(jì)各種機(jī)器和承重結(jié)構(gòu)的裝置能力的未來(lái)工程師。
既分析并設(shè)計(jì)一個(gè)給定結(jié)構(gòu)的涉及應(yīng)力和變形的測(cè)定。這第一章介紹應(yīng)力的概念。
1.2節(jié)所專(zhuān)門(mén)介紹的是靜力學(xué)和他們?cè)谟珊?jiǎn)單結(jié)構(gòu)的成員力量的中心的基本方法連接方面的應(yīng)用。1.3節(jié)將向您介紹應(yīng)力的一個(gè)概念結(jié)構(gòu)的成員，你將看到壓力如何可以從該成員的力量來(lái)決定。在工程分析和設(shè)計(jì)（1.4節(jié)）的一個(gè)簡(jiǎn)短的討論后，你會(huì)陸續(xù)考慮成員在軸向載荷（1.5節(jié)）的正應(yīng)力，造成相等，方向相反的橫向力的應(yīng)用程序（1.6節(jié)）的剪切應(yīng)力，并通過(guò)螺栓和銷(xiāo)的成員的連接所創(chuàng)建的承載壓力（1.7節(jié)）。這些不同的概念將在1.8節(jié)中被應(yīng)用，并對(duì)應(yīng)力的成員進(jìn)行判斷。
有關(guān)靜力學(xué)的簡(jiǎn)短評(píng)述
在本節(jié)中，你將會(huì)學(xué)習(xí)到一種確定簡(jiǎn)單結(jié)構(gòu)各部分受力的基本靜力學(xué)方法。
考慮圖1.1所示的結(jié)構(gòu)，它被設(shè)計(jì)為可以支持30千牛的負(fù)荷的載體。它由一個(gè)用30* 50毫米矩形橫截面的懸臂AB和直徑20毫米的圓形橫截面的桿組成。起重臂和桿通過(guò)在B中的銷(xiāo)連接，并且通過(guò)銷(xiāo)和支架在A和C，分別支承。我們的第一個(gè)步驟應(yīng)該是從其在A和C支持之中分離，再將其表明這些支持在結(jié)構(gòu)（圖1.2）發(fā)揮反應(yīng)繪制結(jié)構(gòu)的自由體圖。需要注意的是該結(jié)構(gòu)的草圖已經(jīng)通過(guò)省略所有不必要的細(xì)節(jié)簡(jiǎn)化。很多人可能在這一點(diǎn)上已經(jīng)認(rèn)識(shí)到，AB和BC是兩個(gè)力的成員。對(duì)于那些你們沒(méi)有，我們將繼續(xù)我們的分析，忽略了這一事實(shí)，并假設(shè)在A和C反應(yīng)的方向是未知的。因此，這些反應(yīng)中的每一個(gè)都將由兩部分組成，Ax和Ay在A和Cx和Cy在C.我們寫(xiě)以下三個(gè)平衡方程來(lái)表示
我們已經(jīng)發(fā)現(xiàn)了兩個(gè)的四個(gè)未知數(shù)，但并不能從這些方程中確定其它兩個(gè)，并且也不能從該結(jié)構(gòu)的自由體圖獲得額外的獨(dú)立方程。我們現(xiàn)在必須拆分結(jié)構(gòu)。考慮到桿AB（圖1.3），我們編寫(xiě)以下平衡方程的自由體圖
將Ay從（1.4）代入到（1.3），我們可以得到Cy=+30千牛。將它表示為在A和C中的矢量形式的反應(yīng)所獲得的結(jié)果，我們有
我們注意到，在A中的反應(yīng)是沿著吊桿AB的軸線定向并且使得壓縮該構(gòu)件。觀察該元件Cx和中C反應(yīng)Cy分別是，正比于從B到C的距離的水平和垂直分量，我們得出結(jié)論，在C反應(yīng)等于50千牛，沿軸線定向桿BC，并導(dǎo)致該成員的張力
這些結(jié)果可能已經(jīng)通過(guò)識(shí)別AB和BC兩種力件，即，是在只有兩點(diǎn)承受力的成員，這些點(diǎn)是A和B構(gòu)件AB和B和C件BC的預(yù)期。的確，對(duì)于兩力構(gòu)件在每個(gè)兩分的作用的力的合力的作用線是相等且相反的并穿過(guò)兩個(gè)點(diǎn)。利用這個(gè)特性，我們可以考慮銷(xiāo)B的自由體圖上針B上的力量是FAB和FC施加的力，分別由成員AB和BC，以及30千牛負(fù)荷（獲得一個(gè)簡(jiǎn)單的解決方案圖1.4A）。我們可以表達(dá)該引腳B是平衡通過(guò)繪制相應(yīng)的力三角形（圖1.4B）。
既然我們已知每個(gè)部件的端部的力，那么我們現(xiàn)在可以確定在這些部件的內(nèi)力。在上一章節(jié)桿BC是一些隨意點(diǎn)D傳遞一個(gè)的部分，我們因此得到兩個(gè)部分BD和CD（圖1.6）。因?yàn)?0千牛的力必須在D點(diǎn)被施加到桿來(lái)保持他們?cè)趦刹糠值钠胶?，我們得出結(jié)論，50千牛的內(nèi)力施加在桿時(shí)產(chǎn)生了一個(gè)30千牛的負(fù)荷，在B處施加時(shí)我們進(jìn)一步檢查從力Fbc和F'bc在圖1的方向入手。 1.6該桿在張力狀態(tài)。類(lèi)似的程序?qū)⑹刮覀兡軌虼_定在桿AB內(nèi)部的力是40千牛，而且桿AB處于壓縮狀態(tài)。
應(yīng)力結(jié)構(gòu)的成員
當(dāng)在前面章節(jié)的內(nèi)容部分中所獲得的結(jié)果表示在給定結(jié)構(gòu)的分析的第一也是必要的步驟，即它們并沒(méi)有告訴我們是否給定的負(fù)載可以安全地支撐。例如棒BC，是否將打破或此載荷向下不向下不僅取決于找到的內(nèi)力Fbc，而且還取決于該桿的橫截面面積和的桿制成的材料中的值。事實(shí)上，內(nèi)力FBC實(shí)際上表示的是分布在該橫截面的整個(gè)面積A基本力的合力（圖1.7）和這些分布式力的平均強(qiáng)度等于在該部的每單位面積/FBCyA的力。無(wú)論桿是否將打破下給定負(fù)載顯然取決于材料的承受分布式內(nèi)力的強(qiáng)度對(duì)應(yīng)的值FBC的能力。因而它取決于力FBC，截面積A，和每單位面積，或分布在桿一個(gè)給定的部分上的力的強(qiáng)度。力的材料，被稱(chēng)為在該部分中的應(yīng)力并由希臘字母S（西格瑪）表示。承受的軸向載荷P是由橫截面面積A的一個(gè)成員的應(yīng)力（圖1.8）除以面積A的負(fù)荷的值P因此得到的。
正號(hào)被用來(lái)表示一個(gè)拉伸應(yīng)力（構(gòu)件在張力）而負(fù)號(hào)來(lái)表示壓縮應(yīng)力（構(gòu)件在壓縮）。
SI公制單位將在該討論中使用，其中P在牛頓（N）和A的平方米（m2）表示，所述應(yīng)力將以牛頓/米(N/m2)來(lái)表示。這個(gè)單位被稱(chēng)為帕斯卡（Pa）。然而，人們發(fā)現(xiàn)，帕斯卡是一個(gè)非常小的量，而且，在實(shí)踐中，必須使用本單位的倍數(shù)，即千帕（kPa）的時(shí)，兆帕，和吉帕斯卡量（GPa）。
當(dāng)使用美國(guó)慣用單位，力P通常以磅（磅）或千磅（千磅），并以平方英寸（英寸2）的橫截面面積A來(lái)表示。應(yīng)力將以磅每平方英寸（psi）或千磅每平方英寸（ksi）的形式來(lái)表達(dá)。
材料力學(xué)的應(yīng)用與設(shè)計(jì)
再次考慮圖1.1的結(jié)構(gòu)，讓我們假定桿BC是由鐵制成的并且最大允許應(yīng)力=165兆帕。請(qǐng)問(wèn)桿 BC桿是否可以安全地支持其將受到的負(fù)荷？之前發(fā)現(xiàn)的在桿FBC所受的力的大小是50千牛，以及桿的直徑為20毫米，我們使用公式（1.5），以確定由給定負(fù)載在桿產(chǎn)生的應(yīng)力。
因?yàn)閷?duì)于s獲得的值是大于所使用的鋼的允許應(yīng)力的s的小，我們的結(jié)論是桿可以安全地支持負(fù)載到其將要進(jìn)行的活動(dòng)。我們給定結(jié)構(gòu)的分析是完整的，也應(yīng)包括在桿AB壓應(yīng)力的中心，以及在銷(xiāo)和軸承產(chǎn)生的應(yīng)力調(diào)查。這將在本章的后面進(jìn)行討論。我們還應(yīng)該確定由給定的載荷所產(chǎn)生的變形是否是可接受的。變形的軸向載荷下的研究將是下章的主題。在受壓構(gòu)件需要額外考慮涉及構(gòu)件的穩(wěn)定性，即，它能夠支持一個(gè)給定的負(fù)荷沒(méi)有經(jīng)歷在結(jié)構(gòu)的突然變化的能力。這也將在下章討論。
工程師的角色并不限于受給定的負(fù)荷條件現(xiàn)有結(jié)構(gòu)和機(jī)器的分析。對(duì)于工程師來(lái)說(shuō)更重要的是新的結(jié)構(gòu)和機(jī)械的設(shè)計(jì)，也就是，相應(yīng)的組件的選擇，以完成執(zhí)行給定任務(wù)。作為設(shè)計(jì)的一個(gè)例子，讓我們回到圖1.1的結(jié)構(gòu)，并且假定該結(jié)構(gòu)是由鋁制成與許用應(yīng)力= 5100兆帕。因?yàn)闂Ubc所受的力仍然會(huì)為給定負(fù)載下p=Fbc=50千牛。我們得出結(jié)論，在由鋁制成的前提下桿的直徑只要26毫米或以上就足夠了。
軸力；正應(yīng)力
我們已經(jīng)指出，在上一節(jié)中考慮的示例的桿BC是一個(gè)兩力構(gòu)件，因此，該桿所受的力FBC和F'bc作用在其端部B和C（圖1.5）沿軸線定向。我們說(shuō)的桿下的軸向載荷。在軸向載荷結(jié)構(gòu)件的實(shí)際例子是，照片1.1所示的橋架。
回到圖1.5所展示的棒bc，我們記得，我們通過(guò)桿傳遞的部分，以確定在桿內(nèi)力和對(duì)應(yīng)的應(yīng)力是垂直于桿的軸線;內(nèi)力因此垂直于部分（圖1.7）的平面與相應(yīng)的應(yīng)力被描述為一個(gè)正常的應(yīng)力。因此，式（1.5）給我們下軸向載荷成員正應(yīng)力。我們還應(yīng)該注意的是，在式（1.5）中，s是通過(guò)將分布在由所述橫截面的面積A的橫截面的內(nèi)力的所得的值P得到的;因此，它代表應(yīng)力在橫截面的橫截面的特定點(diǎn)上的平均值，而不是壓力。在橫截面的給定的點(diǎn)Q定義應(yīng)力，我們應(yīng)該考慮的小區(qū)域DA（圖1.9）。一天劃分DF的大小，我們得到了強(qiáng)調(diào)DA的平均值。讓DA趨近于零，我們?cè)讷@得Q點(diǎn)的應(yīng)力。在一般情況下，在段的一個(gè)給定的點(diǎn)Q的應(yīng)力小號(hào)得到的值是從由通式（1.5）給出的平均應(yīng)力值不同，并且s是在整個(gè)部分而變化。在經(jīng)受相等且相反的集中載荷P和P9（圖1.10A）一細(xì)長(zhǎng)桿，該變化是在一個(gè)部分從集中載荷（圖1.10c）分離開(kāi)來(lái)的，但它明顯是在這點(diǎn)（圖1.10B和d）附近。這意味著，在根據(jù)每個(gè)應(yīng)力面的體積圖。 1.10必須等于所述負(fù)荷的大小P。然而，這是我們可以從我們的靜力學(xué)的知識(shí)推導(dǎo)出，對(duì)于正應(yīng)力在桿的不同部分分配的唯一信息。應(yīng)力在任何給定部分的實(shí)際分配是靜態(tài)的不確定的。要了解更多關(guān)于這種分布，有必要考慮負(fù)載的應(yīng)用在下一章中進(jìn)一步討論的.這個(gè)棒的末端的特定模式所產(chǎn)生的變形。在實(shí)踐中，將假定的正常分布在軸向加載構(gòu)件的應(yīng)力是均勻的，除了在負(fù)載的應(yīng)用的點(diǎn)的附近。應(yīng)力的值則可從式（1.5）來(lái)獲得。但是，我們應(yīng)該認(rèn)識(shí)到，當(dāng)我們假定應(yīng)力的部分，即均勻分布，當(dāng)我們假定內(nèi)力均勻分布在部分分布，從基本的靜態(tài)遵循?，內(nèi)部的合力p必須在段（圖1.11）的質(zhì)心C下施加。這意味著，應(yīng)力的均勻分布是可能的，只有該集中載荷的P動(dòng)作和P9的線通過(guò)考慮（圖1.12）部分的質(zhì)心。這種類(lèi)型的負(fù)荷被稱(chēng)為中心裝載和將被假定為發(fā)生在桁架和銷(xiāo)連接結(jié)構(gòu)，如一個(gè)在圖1.1中視為找到的所有直兩力的成員。然而，如果兩力構(gòu)件被軸向加載，但偏心如圖。 1.13a，我們從圖中所示的構(gòu)件的一部分的平衡的條件下找到。 1.13b在截面的形心和力矩M 5的Pd幾輛M，在一個(gè)給定的部分的內(nèi)力必須等于一個(gè)力P施加的分布力，因此，對(duì)應(yīng)的分布的應(yīng)力是不可能均勻的。應(yīng)力的分布，也不可以是對(duì)稱(chēng)的，如圖1.1所示這一點(diǎn)也將在下一章中進(jìn)行詳細(xì)討論。