過程裝備與控制工程專業(yè)英語翻譯閱讀材料

1、Reading material 1 Static Analysis of BeamsNote that uses up one of the three independent equations of statics, thus only two additional reaction components may be determinated from statics. If more unknow reaction components or moments exist at the support, the problem becomes statically indetermin

2、ate.A bar that is subjected to forces acting trasverse to its axis is called a beam. In this section we consider only a few of the simplest types of beams, such as those shown in Flag.1.2. In every instance it is assumed that the beam has a plane of symmetry that is parallel to the plane of the figu

3、re itself. Thus，the cross section of the beam has a vertical axis of symmetry.Also,it is assumed that the applied loads act in the plane of symmetry,and hence bending of the beam occurs in that plane. Later we will consider a more general kind of bending in which the beam may have an unsymmetrical c

4、ross section.The beam in Fig.1.2（a), with a pin support at one end and a roller support at the other, is called a simply support beam ,or a simple beam. The essential feature of a simple beam is that both ends of the beam may rotate freely during bending, but they cannot translate in lateral directi

5、on. Also ,one end of the beam can move freely in the axial direction (that is, horizontally). The supports of a simple beam may sustain vertical reactions acting either upward or downward .The beam in Flg.1.2(b) which is built-in or fixed at one end and free at the other end, is called a cantilever

6、beam. At the fixed support the beam can neither rotate nor translate, while at the free end it may do both. The third example in the figure shows a beam with an overhang. This beam is simply supported at A and B and has a free at C.Loads on a beam may be concentrated forces, such as P1 and P2 in Fig

7、.1.2(a) and (c), or distributed loads , such as the the load q in Fig.1.2(b),。Distributed loads are characterized by their intensity,which is expressed in units of force per unit distance along the axis of the beam. For a uniformly distributed load, illustrated in Fig.1.2(b),the intensity is constan

8、t; a varying load, on the other hand, is one in which the intensity varies as a function of distance along the axis of the beam.The beams shown in Fig.1.2 are statically determinate because all their reactions can be determined from equations of static equilibrium. For instance ,in the case of the s

9、imple beam supporting the load P1 Fig.1.2(a), both reactions are vertical, and their magnitudes can be found by summing moments about the ends; thus, we find The reactions for the beam with an overhang Fig.1.2 (c)can be found the same mannerFor the cantilever beamFig.1.2(b), the action of the applie

10、d load q is equilibrated by a vertical force RA and a couple MA acting at the fixed support, as shown in the figure. From a summation of forces in vertical direction , we include thatAnd ,from a summation of moments about point A, we findThe reactive moment MA acts counterclockwise as shown in the f

11、igureThe preceding examples illustrate how the reactions(forces and moments) of statically determinate beams may be calculated by statics.The determination of the reactions for statically indeterminate beams requires a considerition of the bending of the beams , and hence this subject will be postpo

12、ned.The idealized support conditions shown in Fig.1.2 are encountered only occasionally in practice. As an example ,long-span beams in bridges sometimes are constructionn with pin and roller supports at the ends. However, in beams of shorter span ,there is usually some restraint against horizonal mo

13、vement of the supports. Under most conditions this restraint has little effect on the action of the beam and can be neglected. However, if the beam is very flexible, and if the horizonal restraints at the ends are very rigid , it may be necessary to consider their effects. Example*Find the reactions

14、 at the supports for a simple beam loaded as shown in fig.1.3(a ). Neglect the weight of the beam.SolutionThe loading of the beam is already given in diagrammatic form. The nature of the supports isexamined next and the unknown components of reactions are boldly indicated on the diagram. The beam ,

15、with the unknow reaction components and all the applied forces, is redrawn in Fig.1.3(b) to deliberately emphasiz this important step in constructing a free-body diagram. At A, two unknow reaction components may exist , since the end is pinned. The reaction at B can only act in a vertical direction

16、since the end is on a roller.The points of application of all forces are carefully noted. After a free-body diagram of the beam is made, the equations of statics are applied to obtain the solution. Note that the concentrated moment applied at C enters only the expressions for summation moments. The

17、positive sign of RB indicates that the direction of RB has been correctly assumed in Fig.1.3(b). The inverse is the case of RAY ,and the vertical reaction at a is downward. Noted that a check on the arithmetical work is available if the calculations are made as shown.閱讀材料1梁的靜力分析在其軸線上受到橫向力的桿我們稱之為梁。在此

18、，我們僅一些最簡(jiǎn)單的類型，如圖1.2所示情形作分析。在每一例中，我們假定有一平面平行于物體表面。從而，橫梁的交叉部分就有一條垂直的軸線。同時(shí)假設(shè)有力作用于平行面上，導(dǎo)致該面發(fā)生彎曲。而后，我們將這種情形推廣至一種更普遍的即橫梁沒有對(duì)稱交叉部分的情況。如圖1.2（a）所示的一端受銷釘支撐、另一端受滾動(dòng)支座支撐的桿我們稱之為“簡(jiǎn)力支撐桿”或“簡(jiǎn)支桿”。簡(jiǎn)支桿的基本特征是其兩端都可自由旋轉(zhuǎn)，但不可橫向移動(dòng)?；蛘邨U的一端可以在軸線方向上（即水平方向）自由移動(dòng)。簡(jiǎn)支桿的支撐端可以維向上或向下的垂直反力。如圖1.2(b)中一端固定，另一端自由的桿稱為“懸梁”。其一端既不可轉(zhuǎn)動(dòng)也不可移動(dòng)，另一端則完全相反。

19、第三個(gè)圖所示為一伸出桿。其在A、B端受支撐，C端自由。梁的負(fù)載有可能是集中力，比如圖1.2（a）1.2（c）中的P1和P2，或是分散力，比如圖1.2（c）中的q。分散力往往通過其密度來表述，單位是在梁的軸向方向上每單位長(zhǎng)度的受力大小。均勻規(guī)律分布的分散負(fù)載，如圖1.2（b）所示，其密度固定不變；而對(duì)于變化的負(fù)載，其密度作為梁的軸向上的函數(shù)而隨時(shí)改變。 圖1.2中的梁均可靜態(tài)確定因?yàn)樗鼈兯艿姆醋饔昧筛鶕?jù)靜態(tài)平衡方程確定。例如，在如圖1.2（a）的簡(jiǎn)支梁的負(fù)載中，其反作用力均為垂直的，其大小均可通過統(tǒng)計(jì)兩端的力矩得出。由此我們得到RA= EQ F(P1(L-a),L) RB= EQ F(P1a,L) 如圖1.2（c）中所示帶有伸出部分的梁的反作用力可以用同樣的方法得出。圖1.2（c）中的懸梁，應(yīng)用負(fù)載在一豎直力RA和一力偶MA的作用下使梁保持平衡。對(duì)豎直方向上的力作統(tǒng)計(jì)，我們得出