过程装备与控制系统工程专业英语

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1、word过程装备与控制工程专业英语学院：化学化工学院1.Static Analysis of Beams 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 everyinstance it is assumed that the beam has a plane of sy

2、mmetry that is parallel to the plane of thefigure 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 moregeneral kind of bend

3、ing in which the beam may have an unsymmetrical cross section. The beam in Fig.1.2, 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 bendin

4、g, but the cannot translate in lateral direction. Also ,one end of the beam can move freely in the axial direction (that is, horizontal). 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 f

5、ree at the other end, is called acantilever beam. At the fixed support the beam can neither rotate nor translate, while at the freeend it may do both. The third example in the figure shows a beam with an overhang. This beam issimply supported at A and B and has a free at C. Loads on a beam may be co

6、ncentrated forces, such as P1 and P2 in Fig.1.2(a) and (c), ordistributed loads loads, such as the the load q in Fig.1.2(b), the intesity. Distributed along the axisof the beam. For a uniformly distributed load, illustrated in Fig.1.2(b),the intensity is constant; avarying load, on the other hand, i

7、s 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 bedetermined from equations of static equilibrium. For instance ,in the case of the simple beam supporting the load P1 Fig.1

8、.2(a), both reactions are vertical, and tehir 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 manner. For the cantilever beamFig.1.2(b), the action of the applied load q is equilibrated by a vertical

9、force RA and a couple MA acting at the fixed support, as shown in the figure. From a summationof forces in certical direction , we include that，And ,from a summation of moments about point A, we find，The reactive moment MA acts counterclockwise as shown in the figure. The preceding examples illustra

10、te how the reactions(forces and moments) of statically determinate beams requires a considerition of the bending of the beams , and hence this subject will be postponed. The idealized support conditions shown in Fig.1.2 are encountered only occasionally in practice. As an example ,long-span beams in

11、 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 movement of the supports. Under most conditions this restraint has little effect on the action of the beam and can be neglected. However

12、, 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 at the supports for a simple beam loaded as shown in fig.1.3(a ). Neglect the weight of the beam. Solution The loading of the beam is

13、 already given in diagrammatic form. The nature of the supports is examined next and the unknow ponents of reactions are boldly indicated on the diagram. The beam , with the unknow reaction ponents and all the applied forces, is redrawn in Fig.1.3(b) to deliberately emphasiz this important step in c

14、onstructing a free-body diagram. At A, two unknow reaction ponents may exist , since 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 abtain the sollution.，RAx=0，2000+100（10）+160（15）RB=0，RB=+27

15、00lb,RAY(20)+2000100（10）160（5）=0，RAY=10lbCheck：，10100160+270=0Note that uses up one of the three independent equations of statics, thus only two additional reaction pones may be determinated from statics. If more unknow reaction ponents or moment exist at the support, the problem bees statically ind

16、eterminate.Note that the concentrated moment applied at C enters only the expressions for summation moments. The 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 c

17、heck on the arithmetical work is available if the caculations are made as shown.横梁的静态分析 一条绕其轴水平放置的棒就是所谓的横梁，本章节我们将研究最简单的横梁模型形式，如图1.2所示。在每个例子中，假设横梁有平整的外形，并且都与其轴线平衡。并且，交叉的部分都有着外形相垂直的关系，同时假设那些负载均作用在平整的外形面上，所以弯曲将会发生在平整的横梁上。稍后我们将研究一个非常普遍常见的并且有垂直于非平整部分的弯曲。 在图1.2（a）的横梁中，横梁的一端由铰链支座支撑着，另外一端由滚动支座支撑，这就是所谓的简单的支撑

18、横梁，或者简单横梁。此简单横梁的特征为两端在弯曲过程可以自由转动，但是不能被转动到侧面其他方向。同时，横梁的一端可以在轴向自由转动（水平方向）。简单横梁的支撑可能会受到向上或者向下的反作用力。 在图1.2（b）中，横梁的一端是固定的，另一端是自由悬空的，这就是所谓的悬空梁。横梁被固定的一端既不能旋转也不能移动，但是另一端则是可以旋转和移动的。图中的第三个例子展示了含有伸出部分的横梁。这个横梁在图A和B中被简单支撑同时在C中有了自由端。 在横梁上的负载可以是垂直的，如图1.2（a）和（c）中的力P1和P2，或者是平均分散的负载，如图1.2（b）中的负载q. 平均分散的负载是以他们的密度分布来区分

19、，这是以在轴线方向上单位长度上得力大小。对于同意的分散负载，如图1.2（b）中所示，其强度为常数；一个变化的负载，在另一方面，是在轴线长度方向上功能强度发生变化的。 在图1.2所示的横梁均为静止确定的，因为它们所有的受力后作用效果均可从静态平衡方程中得到确定。例如，在支撑负载P1的简单横梁所示的情况中，所有的作用效果都是在垂直方向的，并且其大小也可以通过总结受力完成瞬间来确定；而且，我们可知：同理我们可以找出确定图1.2（c）中所示的有外伸端的横梁作用效果。 对于悬梁图1.2(b)，如图中所示，应用负载q的作用效果可以被固定端的垂直力RA和力矩MA抵消平衡。从垂直方向上力的总结可得，并且通过对

20、质点A的瞬间的总结可得MA的工作效果如图所示，是延时钟方向。 先前图示的例子说明静态确定横梁的作用效果可以从方程中计算出来。静态非确定横梁的作用效果的确定需要考虑横梁的形变效果，这种研究将在以后的课程学习中讨论。 图1.2中所示的理想化支撑条件只是偶尔在实际中才会遇到。举例说明如，在桥的大跨度横梁两端有时候也被建成用铰和滚动支撑。当然，在短一点的横梁上，经常会有对支撑的水平移动的制约力。在大部分情况下，这种制约对横梁作用效果有很小的影响，是可以被忽略的。当然，如果横梁是非常容易弯曲的，并且假如两端水平的制约力是非常有效果的，那就有必要考虑它们共同的作用效果。 例题 找出图1.3（a）中所示的简

21、单横梁受力下的反作用力，忽略横梁自身的重量。 解答 横梁所受到的力已经在图示中给出。支撑力的性质接下来就会测出，并且这些部分中的未知组成部分被大胆的体现在图示中。有未知反作用力组成部分和所有已提供力的横梁被重新展示在图1.3（b）中，来刻意的强调构建这个自由体图示的重要步骤。在A中，由于一端是被别住的，或许存在两个未知的反作用力。由于B端是在滚动作用上，所以其反作用力只可能在垂直作用力上。所有力的作用点都被认真标记出。当一自由体的受力图被完成之后，需要有静态方程来求解，RAx=0，2000+100（10）+160（15）RB=0，RB=+2700lb,RAY(20)+2000100（10）16

22、0（5）=0，RAY=10lb验证：，10100160+270=0 应注意包含了三个非独立静态方程中的一个，所以只有两个附加的反应力组成部分可能从方程中被求出。假如有更多的反应力的组成部分或是瞬间存在于支撑中，问题成了静态不可解性的。 注意在点c中心的力集中的瞬间只是出现在所有瞬间总和的表现中。RB正向标记说明RB已经在图1.3（b）中被正确的假定。相反的是RAy的情况，并且A点的垂直反作用力是向下的，注意假如计算如上所示，那么计算过程中的验证是有效的。2.Share Fore and Bending Moment in BeamsLet us now consider, as an exam

23、ple, a cantilever beam acted upon by an inclined load P at its free end Fig.1.5 (a).If we cut through the beam at a cross section mn and isolate the left-hand part of the beam as free body Fig.1.5 (b), we see that the action of the removed part of the beam (that is, the right-hand part) upon the lef

24、t part must be such as to hold the left-hand part in equilibrium. The distribution of stresses over the cross section mn is not known at this stage in our study，but we do know that the resultant of these stresses must be such as to equilibrate the load P. It convenient to resolve the resultant into

25、an axial force N acting normal to cross section and passing through the centroid of the cross section ,a shear force V acting parallel to the cross section, and a bending moment M acting in the plane of the beam.The axial force, shear force, and bending moment acting at a cross section of a beam are

26、 known as stress resultants. For a statically determinate beam, the stress resultants can be determined from equations of equilibrium. Thus, for the cantilever beam pictured in Fig.1.5, we may write three equations of statics for the free-body diagram shown in the second part of the figure. From sum

27、mations of forces in the horizontal and vertical directions we find, respectively，N=Pcos V=Psinand, from a summation of moments about an axis though the centroid of cross section mn, we obtain M=PxsinWhere x is the distance from the free end to section mn,。Thus, through the use of a free-body diagra

28、m and equations of static equilibrium we are able to calculate the stress resultants without difficulty. The stresses in the beam due to the axial fore N acting alone have been discussed in the text of Uint.2; Now we will see how to obtain the stresses associated with bending moment M and the shear

29、force V.The stress resultants N ,V and M will be assumed to be positive when they act in the directions shown in Fig.1.5b.This sign convention is only useful, however ,when we are discussing the equilibrium of the left-hand part of the beam is considered, we will find that the stress resultants have

30、 the same magnitudes but opposite directionssee Fig.1.5(c).Therefore, we must recognize that the algebraic sign of a stress resultant does not depend upon its direction in space ,such as to the left or to the right ,but rather it depends upon its direction with respect to the material against which

31、it acts .To illustrate this fact, the sign conventions for N,V and M are repeated in Fig.1.6, where the stress resultants are shown acting on an element of the beam.We see that a positive axial force is directed away from the surface upon which it acts (tension), a positive shear force acts clockwis

32、e about the surface upon which it acts, and a positive bending moment is one that presses the upper part of the beam.Example A simple beam AB carries two load, a concentrated force P and a couple Mo, acting as shown in Fig.1.7（a）. Find the shear force and bending moment in the beam at cross sections

33、 located as follows: (a) a small distance to the left of the middle of the beam and (b) a small distance to the right of the middle of the beam.SolutionThe first step in the analysis of this beam is to find the reaction Ra and Rb .Taking moment about ends and A and B give two equations of equilibriu

34、m, from which we findNext, the beam is cut at a cross section just to the middle, and a free-body diagram is drawn of either half of the beam. In this example we choose the left-hand half of the beam, and the corresponding diagram is shown in Fig.1.7(b).The force P and the reaction Ra appear in this

35、 diagram , as also do the unknown shear force V and bending moment M, both of which are shown in their positive directions, The couple Mo does not appear in the figure because the is cut to the left of the point where Mo is applied. A summation of force in the vertical direction gives V=Which shows

36、that the share force is negative; it acts in the opposite direction to that assumed in Fig.1.7(b) .Taking moments about an axis through the section where the beam is cut Fig.1.7(b) gives M=Depending upon the relative magnitudes of the terms in this equation, we see that the bending moment M may be e

37、ither positive or negative.To obtain the stress resultants at a cross section just to the right of the middle, we cut the beam at that section and again draw an appropriate free-body diagram Fig.1.7(c). The only difference between this diagram and the former one is that the couple Mo now acts on the

38、 part of the beam to the left of the cut section. Again summing force in the vertical direction, and also taking moments about an axis through the cut section, we obtain V= M=We see from these results that the shear force does not change when the cut section is shifted from left to right of the coup

39、le Mo, but the bending moment increases algebraically by an amount equal to Mo.梁的剪力和弯矩让我们来研究，一个例子，自由端上作用有倾斜负载P的悬臂梁，如图（1.5a）所示。如果我们在截面mn处切断并将梁的左端隔离作为一个自由体，如图（1.5b）所示，我们看到，被隔离出来的那部分横梁（即右半部分）必须和左半部分一样是处于平衡状态。在截面mn上的应力分布，在我们现阶段的学习中是不可知道的，但是我们可以知道应力的合力一定与负载P平衡。这样我们就很方便去解决由此而产生的一个轴向力N作用于截面和通过轴心的横截面，平行于截面的

40、剪力V和作用在梁的平面上的弯矩M。作用在梁截面的轴向力，剪力和弯矩都是可知道的力。对于一个静态梁，力可以由平衡方程来决定。因此，对于如图（1.5）的悬臂梁，我们可以写出三个关于自由体图解的静态方程，如图第二部分所示。水平和垂直方向的合力，我们可以发现，分别地为，N=Pcos V=Psin通过mn截面形心的轴的弯矩和为M=PxsinX为自由端到截面的距离。因此，通过运用自由端的图解和静态平衡方程，我们可以很容易地计算出力。由梁上轴向力N单独作用引起的应力在第二单元的章节已经讨论过。现在我们需如何获得与应力相关的弯矩M和剪力V。我们假定力N、V和M的方向是正向的，它们的作用方向如图1.5b所示。这

41、一规定是有用的，但是，只能是在我们在讨论梁左端的平衡时。如果讨论梁的右端，我们将会发现力有相同的作用效果，但是方向是相反的（看图1.5b）。因此，我们必须承认力的代数和是不依赖于它的空间方向的，如向左或者是向右，而是依赖于作用在材料上的反方向。为了说明事实，对N、V、M的规定在图1.6有复述，即为力作用于横梁上产生的效果。我们发现正面的的轴力反作用于其表面（拉力），正向剪力顺时针作用于表面，以及正面的弯矩是作用在压缩梁上部的正向弯矩之一。例题 一个简单梁AB有着两个负载，一个集中负荷P和一个力偶Mo，作用如图1.7a所示。找出梁上截面处的的剪力和弯矩，如下：（a）梁中间到左端的一小段距离(b)

42、梁中点到右端的一小段距离。解答 这个梁的分析的第一步是找出力Ra和Rb。取A、B端可列出两个平衡方程，我们发现，接下来，将梁在截面出剪断并仅取中点的左端，将梁的一半作为自由体做图解。在这个例子中，我们选择梁的左半边为研究对象，其相应的图解如1.7b所示。负荷P和力Ra出现在这个图解上，同样也有未知的剪力V和弯矩M，它们都是为正方向的，力偶则不出现在这个图上，因为梁被切断点的左端没有Mo作用。在垂直方向的合力为V=这说明剪力是负的；今后，假定其作用在反方向上，如图1.7b。取通过分截面对轴的矩，如图1.7b所示，则有 M=在这个公式中，M依赖于相关的代数和，我们可以看到M可能是负的或者是正的。为

43、了获得截面的力，仅取梁的右端为研究对象，我们在梁的截面处切断，然后重新画自由体图解1.7c。这个图解跟前面那个的不同点仅仅是现在力偶Mo作用于切断面的左端上。重新求垂直方向上的合力，同时求切断面对轴的弯矩，我们可以得到 V= M=我们从这些结果可以看到，当切断面由力偶Mo左端转换到右端，剪力都不会改变，但是弯矩的代数和增量等于Mo。3.Theories of strength1. Principal stresses The state of the stress at a point in a structural member under a plex system of loading

44、is described by the magnitude and direction of the principal stresses. The principal stresses are the maximum values of the normal stresses at the point; which act on the planes on which the shear stress is zero. In a two-dimensional stress system, Fig.1.11, the principal stresses at any pint are re

45、lated to the normal stress in the x and y directions x and y and the shear stress xy at the point by the following equation:Principal stresses,The maximum shear stress at the point is equal to half the algebraic between the principalstresses:Maximum shear stress,pressive stresses are conventionally

46、taken as negative; tensile as positive.2. Classification of pressure vessels For the purpose of design and analysis, pressure vessels are sub-divided into two classes depending on the ratio of the wall thickness to vessel diameter: thin-wall vessels, with a thickness ratio of less than 1/10, and thi

47、ck-walled above this ratio. The principal stresses acting at a point in the wall of a vessel, due to a pressure load, are shown in Fig.1.12. If the wall is thin, the radial stress 3 will be small and can be neglected in parison with the other stresses , and the longitudinal and circumferential stres

48、ses1 and2 can be taken as constant over the wall thickness. In a thick wall, the magnitude of the radial stress will be significant, and the circumferential stress will vary across the wall. The majority of the vessels used in the chemical and allied industries are classified as thin-walled vessels.

49、 Thick-walled vessels are used for high pressures.3. Allowable stress In the first two sections of this unit equations were developed for finding the normal stress and average shear stress in a structural member. These equations can also be used to select the size of a member if the members strength

50、 is known. The strength of a material can be defined in several ways, depending on the material and the environment in which it is to be used. One definition is the ultimate strength or stress. Ultimate strength of a material will rupture when subjected to a purely axial load. This property is deter

51、mined from a tensile test of the material. This is a laboratory test of an accurately prepared specimen, which usually is conducted on a universal testing machine. The load is applied slowly and is continuously monitored. The ultimate stress or strength is the maximum load divided by the original cr

52、oss-sectional area. The ultimate strength for most engineering materials has been accurately determined and is readily available. If a member is loaded beyond its ultimate strength it will fail-rupture. In the most engineering structures it is desirable that the structure not fail. Thus design is ba

53、sed on some lower value called allowable stress or design stress. If, for example, a certain steel is known to have an ultimate strength of 110000 psi, a lower allowable stress would be used for design, say 55000 psi. this allowable stress would allow only half the load the ultimate strength would a

54、llow. The ratio of the ultimate strength to the allowable stress is known as the factor of safety:We use S for strength or allowable and for the actual stress in material. In a design:SAThis so-called factor of safety covers a multitude of sins. It includes such factors as the uncertainty of the loa

55、d, the uncertainty of the material properties and the inaccuracy of the stress analysis. It could more accurately be called a factor of ignorance! In general, the more accurate, extensive, and expensive the analysis, the lower the factor of safety necessary.4. Theories of failureThe failure of a sim

56、ple structural element under unidirectional stress (tensile or pressive) is easy to relate to the tensile strength of the material, as determined in a standard tensile test, but for ponents subjected to bined stresses (normal and shear stress) the position is not so simple, and several theories of f

57、ailure have been proposed. The three theories most monly used are described below:Maximum principal stress theory: which postulates that a member will fail when one of the principal stresses reaches the failure value in simple tension,e. The failure point in a simple tension is taken as the yield-po

58、int stress, or the tensile strength of the material divided by a suitable factor of safety. Maximum shear stress theory: which postulates that failure will occur in a plex stress system when the maximum shear stresses reaches the value of the shear stress at failure in simple tension.For a system of

59、 bined stresses there are three shear stresses maxima:， （1.10）In the tensile test, (1.11)The maximum shear stress will depend on the sign of the principal stresses as well as their magnitude, and in a two-dimensional stress system, such as that in the wall of a thin-walled pressure vessel, the maxim

60、um value of the shear stress may be given by putting 3 =0 in equations 1.10. The maximum shear stresses theory is often called Trescas, or Guests theory.Maximum strain energy theory: which postulates the failure will occur in a plex stress system when the total strain energy per unit volume reaches

61、the value at which failure occurs in simple tensile.The maximum shear-stress theory has been found to be suitable for predicting the failure of ductile material under plex loading and is the criterion normally used in the pressure-vessel design.力学理论1. 主应力在一个错综复杂的系统的结构中，某点的受力情况通常由主应力的方向和大小来描述。主应力是质点所

62、受到的正应力的最大有效值；它作用于抗剪应力为零的平面上。在一个二维的力学系统中，如图1.11所示，任何质点的主应力都与在X和Y方向上的应力x和y有关，同时该质点的抗剪应力 xy 可由下列公式确定： 主应力 该点的最大抗剪应力是主应力的代数差的一半：最大抗剪应力，max=（12）压缩应力是常见的，但不是很明显，而拉伸应力则是明显的。2压力容器的分类 为了设计和分析，压力容器细分为两类，依据于厚壁和容器直径的比率：薄壁容器，比率小于1/10，同时厚壁容器的比率超过1/10。 作用于容器壁上某点的主要压力源于压力负载，如图1.22所示。假如壁是薄的，放射状的压力3将会很小，并且通过与其他压力的比较是

63、可以忽略的，而纵向的和圆周上的压力1和2就会被视为与壁厚无关的常数。在薄壁上，放射性压力的大小将会是很重要的，而圆周的压力将会绕容器壁变化。应用于化工和联合工业的主要压力容器被分为薄壁容器。厚壁容器主要应用于高压环境下。3. 允许应力本章节开始的两个部分中，方程是为了找到结构中的主要压力和平均切向压力而产生的。这些方程同样在结构力已知的情况下，可以被用于选择结构组成。材料强度可以从以下几方面来定义，依据材料自身和其使用环境。一种定义是其最大的强度和压力。最大强度是指当其受到纯轴向负载作用下发生断裂的压力。这项属性将会从材料的拉伸试验中测出。这是在一个具备普遍检测能力的机器上对某一准确制备的样品

64、进行检测的实验室试验。负载缓慢增加，并且是始终处于监视之下。最大强度或压力是可以在原始交汇区域分开的最大负载。对于大多数的设计材料来讲，其最大强度是将会被准确测出并给出。假如某一部分承受的载荷超过了他的最大强度值，它将断裂。在大多数的设计结构中人们不希望结构崩溃。所以，设计是基于更小的有效值，它通常被称作允许压力或设计压力。举个例子，假如一个确定的钢坯是已知有最大强度值110000每平方英寸磅，一个低一点的强度将会被用于其设计，比如55000每平方英寸磅，这个允许强度将会承认最大强度的一半。最大强度值与允许强度值的比值被称作安全系数：或者 我们用S表示最大强度或者最大允许压力，用表示材料受到的实际负载，即SA这个所谓的安全系数包括了很多不确定性。其中包括负载的不确定性，材料性能的不确定性，以及压力分析的不准确度等因素。它可以更准确的被称为忽略系数！事实上，分析得更准确、更广泛