过程装备与控制工程专业英语上册

1、 过程装备与控制工程专业英语翻译作业 龙飞啸天Reading Material 1 Static Analysis of BeamsA bar that is subjected to forces acting transverse to its axis is axis is called a beam. In this section we will consider only a few of the simplest types of beams, such as those shown in Fig.1.2. In every instance it is assumed that

2、 the beam has a plane of symmetry that is parallel to the plane of the figure 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 consid

3、er a more general kind of bending in which the beam may have an unsymmetrical cross 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 supported beam, or a simply beam . The essential feature of a simple beam is that both ends of the b

4、eam may rotate freely during bending, but they cannot translate in the lateral direction. 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 and downward. The beam in Fig.1.2(b) w

5、hich is build-in or fixed at one end and free at the other end , is called a cantilever 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. The beam is simply supported at A and

6、B and has a free end at C. Loads on a beam may be concentrated forces, such as P1 and P2 in Fig.1.2(a) and (c). Distributed loads, such as 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 be

7、am. For a uniformly distributed load, illustrated in Fig.1.2(b),the intensity is constant; 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 beans shown in Fig.1.2 are statically determinate because all their reactions ca

8、n be determined from equations of static equilibrium.For instance,in the case of simple beam supporting the load P1Fig.1.2,both reactions are vertical,and their magnitudes can be found by summing moments about the ends;thus,we find Ra=P1(L-a)/L Rb=P1a/LThe reactions for the beam with an overhang Fig

9、.1.2(c)can be found in the same manner. For the cantilever beam Fig.1.2(b),the action of the applied 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 the vertical direction ,we conclude that Ra=qbAnd, from

10、 a summation of moments about point A, we find Ma=qb(a+b/2)The reactive moment Ma acts counterclockwise as shown in the figure. The preceding examples illustrate how the reactions (forces and moments) of statically determinate beams many be calculated by statics. The determination of the reactions f

11、or statically indeterminate beams requires a consideration of the bending of the beams ,and hence this subject will be postponed. The preceding examples illustrate how the reaction (forces and moments) of statically determinate beams may be calculated by statics. The determination of the reactions f

12、or statically indeterminate beams requires a consideration 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 bridges sometimes are constructed with p

13、in and roller supports at the ends. However, in beams of shorter span, there is usually some restraint against horizontal movement of the supports. Under most conditions this restraint has little effect on the beam and can be neglected. However, if the beam is very flexible , and if the horizontal r

14、estraints at the ends are very rigid, it may be necessary to consider their effects.Example Find the reactions at the supports for a simple vbeam loaded as shown in Fig.1.3(a). Neglect the weight of the beam.Solution The loading of the beam is already given in diagrammatic form. The nature of the su

15、pports is examined next and the unknown components of these reactions are boldly indicated on the diagram. The beam, with the unknown reaction components and all the applied forces, is redrawn in Fig.1.3(b) to deliberately emphasize this important step in constructing a free-body diagram. At A, two

16、unknown reaction components may exist, since the end is pinned. The reaction at B can only act in a vertical direction 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 applied to obtain

17、the solution. Fx=0,RAx=0 MA=0+,2000+100(10)+160(15)-RB(20)=0, RB=+2700 lb MB=0+,y=0+,-10-100-160+270=0 Note thatFx=0 uses up one of the three independent equations of statics, thus only two additional reaction components may be determined from statics. If more unknown reaction components or moments

18、exist at the support, the problems become statically indeterminate. Note that the concentrated moment applied at C enters only into the expressions for the summation of moments. The positive of RB indicates that the direction of RB has been correctly assumed in fig.1.3(b). The inverse is the case of

19、 Ray, and the vertical reaction at A is downward. Note that a check on the arithmetical work is available if the calculations are made as shown.梁的静态分析梁是指一根在轴线上受到横向作用力的杆。在这篇课文中我们仅仅研究的只是一些简单常见的梁，比如图12中所示的那些。每个例子中梁都被假设有一个与其本身所在平面平行的对称平面。因此，梁在横断面有一个垂直轴向对称面。与此同时，假定负载只在对称的面中起作用，因此梁的弯曲部分仅在其中的一个平面中发生。以后我们可能

20、会讨论更多具有普遍性的弯曲发生在不对称的横断面弯曲的梁。图1.2（a）示的梁，一端铰链支撑,另一方则以滚动轴承的支持,这就是所谓的简支梁支架顶梁,或梁。主要特点:梁弯曲的面内自由旋转的,但它们不可能水平移动。此外,梁的结束的轴向上可以自由移动。简支梁支架可以受到垂直方向的反作用力,垂直向上或者向下。如图1.2（b）中的梁，一端是嵌在固定端,另一端是自由端的。所以我们都叫悬臂梁。固定端梁也允许转动不允许移动,但这章反映到第三个例子。这是一种混合梁,梁是在A、B各有支点支撑,而c端是自由端。作用在梁上的载荷可能是集中力，如图1.2（a）和（c）中的P1和P2，也可能是分布载荷，图1.2（b）中的力

21、q。分布载荷的特点就是使梁在其轴线单位距离上受到特定的力。对于一个均匀分布的载荷，其力的强度是一个常数，比如1.2（b）中所示;另一方面，一个变化的负载，其力的强度随着梁长而改变。图1.2中梁是静止的，因为它们所有反作用力可以通过平衡方程求得。举个例子也就是说，在简单梁的情况下负载P1【图1.2（a）】和它的反作用力，两者的方向是垂直的，而且它们的大小可以通过弯矩计算，因此，我们得到： RAP1(L-a)/L, RB=P1a/L 同理可以确定悬臂梁上的发作用力。对于悬臂梁【图1.2（b）】，载荷q的作用力被一个垂直的力RA和一个在固定端起作用的力偶MA平衡，如图所示，从垂直方向上计算力的总和，

22、我们可以得出结论：RAqb并且求A点的矩，我们发现:MAqb（ab/2） 如图中显示作用的矩MA是逆时针方向的。 前述的例子说明静定梁的反作用（力和矩）可以通过静力学的原理计算出来。确定静不定梁的应力要考虑梁的挠度，因此，还需要了解更多方面的知识。 在图1.2中所示的理想条件支座实际中只有在特殊场合才遇见。举个例子说，长梁的端点有时是把绞支座和滚动支座连在一起的。而在较短跨度的梁中，通常梁的水平条件受到外在条件限制。在多数情况下这抑制在梁的作用上完全不起作用，所以可以忽略。然而，如果梁容易弯曲，而且端点水平的限制力非常大，就必须考虑它们带来的影响了。 例子：计算在图1.3（a）中的支持简单梁的

23、作用力，梁的重力不计。 解： 梁的示意图以图标形式给出。架的性质以后会被检验，未知的反作用力已经在图上明显的标出。图1.3（b）中未知数和外加力的梁，主要表示一个自由体受力分析图。在A点存在两个未知的反作用力，因为它的端点是铰支座。因为B端固定在一个滚子上，梁在B端只受一个垂直方向的反作用力。在分析梁的受力图后，静力学的平衡方程就可以列出来了： Fx=0, RAx=0 MA=0+,2000+100(10)+160(15)-RB(20)=0,Rb=+2700 lb MB=0+,RAy(20)+2000-100(10)-160(5)=0,RAy=-10 lb 验证:Fy=0+,-10-100-16

24、0+270=0注:Fx= 0这是其中的三个静力学方程,因此,从静力学的角度仅能确定两个未知的压力。虽然许多物体的支持的的作用、问题变成拉压超静不定问题。注意C点应用扭矩仅仅表示集中力矩之和。RB用力方向图13(b)假设正确。剪应力的下方。值得注意的是,这可校核计算的结果是合理的。Reading Material 2Shear Force and Bending Moment in Beams Let us now consider, as an example, a cantilever beam acted upon bu an inclined load P at its free end

25、 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 left-hand part must be such as to hold the left-hand part in equilibrium.

26、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 is convenient to resolve the resultant into an axial force N acting normal to the cross section and passi

27、ng 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 known as stress resultants, For a statically determinat

28、e 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 summations of forces in the horizontal and vertical direct

29、ions we find, respectively, N=Pcos V=Psinand, from a summation of moments about an axis through the centroid of cross section mn, we obtainM=Pxsinwhere x is the distance from the free end to section mn. Thus , through the use of a free-body diagram and equations of static equilibrium, we are able to

30、 calculate the stress resultants without difficulty. The stresses in the beam due to the axial force N acting alone have been discussed in the text of Unit.2; Now we will see how to obtain the stresses associated with bending moment M and the shear force V. The stress resultants N,V and M will be as

31、sumed to be positive when they act in the directions shown in Fig.1.5(b). This sign convention is only useful, however, when we are discussing the equilibrium of the left-hand part of the beam. If the right-hand part of the beam is considered, we will find that the stress resultants have the same ma

32、gnitudes but opposite directions see Fig.1.5(c). Therefore, we must recognize that the algebraic sign of a stress resultants 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 it acts.

33、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 clockwise abou

34、t the surface upon which it acts, and a positive bending moment is one that compresses the upper part of the beam.Example A simple beam AB carries two loads, a concentrated force and a couple M, acting as shown in Fig.1.7(a). Find the shear force and bending moment in the beam at cross sections loca

35、ted 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 middle of the beam.Solution The first step in the analysis of this beam is to find the reactions Ra and Rb. Taking moments about ends A and B gives two equations of eq

36、uilibrium, from which we findRa=3P/4-Mo/l Rb=P/4+Mo/lNext, the beam is cut at a cross section just to the left of 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, as also do the unknown s

37、hear force V and bending moment M, both of which are shown in the figure because the beam is cut to the left of the point where Mo is applied. A summation of forces in the vertical direction givesV=Ra-P=-P/4-Mo/Lwhich shows that the shear force is negative; hence, it acts in the opposite direction t

38、o that assumed in Fig.1.7(b). taking moments about an axis through the cross section where the beam is cut Fig.1.7(b) gives M=RaL/2-PL/4=PL/8-MO/2Depending upon the relative magnitudes of the terms in this equation, we see that the bending moment M may be either positive or negative. To obtain the s

39、tress 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 couple Mo now acts on the part of the beam to the left of the cut section. Again summing f

40、orces in the vertical direction, and also taking moments about an axis through the cut section, we obtainV= -P/4-Mo/L M=PL/8+Mo/2We see from these results that the shear force does not change when the cut section is shifted from left to right of the couple Mo, but the bending moment increases algebr

41、aically by an amount equal to Mo. 梁的剪力与弯矩现在让我们来考虑一下, 一端受载荷P倾斜的悬臂梁,如图1.5(a)。假如说，我们在梁上取截面mn拿左边自由端梁体分析，如图1.5（b），我们得到，远离左边部分的梁（即右端的部分）一定要靠左边部分支撑才能达到平衡的。在横断面mn上的应力分布不是现阶段学习可以知道的，但是我们都知道，这些应力的合力一定与负载P平衡。我们可以用这样简单的方法解决问题，把合力分解为平行轴向的力N,平行于截面的剪力,弯矩M作用在梁上。轴向力、剪切,横梁弯矩都可以看作是对中心已知的应力是合理的。为一个静态的梁、应力是合理的、平衡方程的决心。因

42、此,图的分析,我们对于图1.5悬臂梁在水平方向和垂直方向的并列三静力学方程出发,如下: NPcos VPsin对于截面mn中心力矩，我们得： MPxsin这里x是mn到自由端的距离。于是通过自由端受力的情况和静态平衡方程分析，很容易计算合应力大小。梁的应力已经在第二单元中独立讨论过。现在讨论获得弯矩M和剪力V。图1.5（b）中所示，合剪力N、V和合力矩M的方向设定为正。当我们在讨论右边的梁时，这个假定是有正确的。讨论右边时，我们发现左右两应力合量大小相等，但方向相反如图1.5（c）。所以，我们可以认识到应力合量的代数和并不是取决于应力的空间方向，而是取决于力的作用效果。为表明此道理，N、V、M

43、的符合规定将会在图1.6中再次出现，梁的应力对梁的作用效果将会显现出来。我们看到一个正向轴应力的方向远离梁的表面（张力），剪力通常沿顺时针方向，而弯矩沿轴线压缩。 例如：一个简单梁受两个载荷，一个集中力P和一个力偶Mo，如图1.7（a）所示。找到梁截面上的剪力和弯矩如下：（a）中央靠左微小距离处（b）中央靠右微小距离处。方法：第一步分析受力，找出反作用力RA和RB。分别以A点和B点做矩，列出方程，从中发现：RA=3P/4-Mo/L RB=P/4+Mo/L下一步把梁从中央靠左方截断，并分析任意半段。在此例子中，我们选左半部分的梁作为研究对象，对应受力如图1.7（b）所示。力P和反力RA出现在这个

44、示意图中，而且包括未知剪力V和弯矩M，令它们均为正向，力偶Mo没有出现在这个图中，因为这部分梁是Mo起作用的梁的一部分。由截面垂直方向可得: V=RA-P=-P/4-Mo/L这说明剪切是负的,因此,它真正的方向与图形的假设相反的方向。梁柱截面的轴,柱弯矩方程式,可以利用: M=RA L/2-PL/4=P L/8-Mo/2依据这个方程会产生大小关系，我们知道结果可能是正的也可能是负的。 得到的右边中心地段的内力、我们在这部分只是画出相应的调整方案。右边部分,一节左边和唯一的区别是力偶Mo梁上的一部分,作用横梁上。再一次计算,我们得到: V=-P/L-Mo/L M=PL/8+Mo/2我们看到，当截

45、面从左端到右端剪力并没有改变，但是弯矩增加了一个力偶Mo的代数和。Reading Material 3Theories of Strength1. principal stressesthe state stress at a point in a structural member under a complex system of loading is described by the magnitude and direction of the principal stressed. The principal stressed are shear stress is zero.In a

46、 two-dimensional stress system ,Fig.1.11,the principal stresses at any point are related to the normal stresses in the x and y directionsx and y snd the shear stress xy at the point by the following equation: The maximum shear stress at the point is equal to half the algebraic difference between the

47、 principal stresses: Compressive stresses are conventionally taken as negative ;tensile as positive .2. classification of pressure vesselsFor the purposes of design and analysis , pressure vessels are sub-diveded into two classes dipending on the ratio of the wall thickness to wessel diamerer:thin-w

48、alled vessels ,with a thickness ratio of less than 1/10 ,and thick-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.12. if the wall is thin t he radial stress 3 will be smell and can be stresses 1 and 2 can be taken

49、 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 are used for high pre

50、ssures.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 is known . The strength of a material ca

51、n be used .One definition is the ultimate strength or stress. Ultimate strength is the stress at which a material will rupture when subjected to a purely axial load .This property is determined from a tensiletest of the material .This a laboratory test of an accurately preared specimen which usually

52、 is conducted on a universal testing machine .The load is maximum load divided by the original cross-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. I

53、n most engineering structures it is desirsble that the structure not fail .Thus design is based onsome lower value called allowable stress or design stress. If , for example , a certain steel is known to have an ultimate strength of psi, a lower allowable stress would be used for design ,say 55000ps

54、i . This allowable stress would allow only half the load the ultimate strength would allow.The ratio of the ultimate strength to the allowable stress is known as the factor of safety: We use S for strength or allowable stress and for actual stress in a material .In a design:This so-called facor of s

55、afety covers a multitude of sins . It includes such factors as the uncertaintu of the load , the uncertainty of the material properties , and the inaccuracy of the stress analusis . It could more accurately be called a factor of ignorance! In general , the more accurate,extensive,and ecpensive the a

56、nalysis , the lower the factor of safety necessary.4. Theories of Failure The failure of a simple structural element under unidirectional stress (tensile or compressive ) is easy to relate to the tensile strength of the material , as determined in a standard tensiletest,but for components subjected

57、to combined stresses (normal and shear stress ) the pisition is not so simple , and seceral theories of failure have been proposed . The three theories most commonly used are described below: Maximmum 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-point stress , or the tensile stress at failure in simple tension. For a sustem of combined stresses there are three shear stresses macima: In