過程裝備與控制工程專業(yè)英語

1、過程裝備與控制工程專業(yè)英語學院：化學化工學院1.static analysis of beams1 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 symm

2、etry 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 consider a more general kind of bend

3、ing in which the beam may have an unsymmetrical cross section.2 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 bendi

4、ng, 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 .3 the beam in flg.1.2(b) which is built-in or fixed at one end and

5、 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. this beam is simply supported at a and b and has a free at c.4 loads on a beam may

6、 be concentrated forces, such as p1 and p2 in fig.1.2(a) and (c), or distributed loads loads, such as the the load q in fig.1.2(b), the intesity. distributed along the axis of the beam. for a uniformly distributed load, illustrated in fig.1.2(b),the intensity is constant; a varying load, on the othe

7、r hand, is one in which the intensity varies as a function of distance along the axis of the beam.5 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 simple beam supporting the lo

8、ad p1 fig.1.2(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.6 for the cantilever beamfig.1.2(b), the action of the applied load q is equilibrated b

9、y a vertical force ra and a couple ma acting at the fixed support, as shown in the figure. from a summation of 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.7 the preceding e

10、xamples illustrate 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.8 the idealized support conditions shown in fig.1.2 are encountered only occasionally in practice. as an example ,l

11、ong-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 movement of the supports. under most conditions this restraint has little effect on the action of the beam and can be n

12、eglected. 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.9 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.10 solution the lo

13、ading of the beam is already given in diagrammatic form. the nature of the supports is examined next and the unknow components of reactions are boldly indicated on the diagram. the beam , with the unknow reaction components and all the applied forces, is redrawn in fig.1.3(b) to deliberately emphasi

14、z this important step in constructing a free-body diagram. at a, two unknow reaction components 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=0，2

15、000+100（10）+160（15）rb=0，rb=+2700lb,ray(20)+2000100（10）160（5）=0，ray=10lbcheck：，10100160+270=0 note that uses up one of the three independent equations of statics, thus only two additional reaction compones may be determinated from statics. if more unknow reaction components or moment exist at the sup

16、port, the problem becomes statically indeterminate. 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

17、reaction at a is downward. noted that a check on the arithmetical work is available if the caculations are made as shown.橫梁的靜態(tài)分析1 一條繞其軸水平放置的棒就是所謂的橫梁，本章節(jié)我們將研究最簡單的橫梁模型形式，如圖1.2所示。在每個例子中，假設(shè)橫梁有平整的外形，并且都與其軸線平衡。并且，交叉的部分都有著外形相垂直的關(guān)系，同時假設(shè)那些負載均作用在平整的外形面上，所以彎曲將會發(fā)生在平整的橫梁上。稍后我們將研究一個非常普遍常見的并且有垂直于非平整部分的彎曲。2 在圖1.2（a

18、）的橫梁中，橫梁的一端由鉸鏈支座支撐著，另外一端由滾動支座支撐，這就是所謂的簡單的支撐橫梁，或者簡單橫梁。此簡單橫梁的特征為兩端在彎曲過程可以自由轉(zhuǎn)動，但是不能被轉(zhuǎn)動到側(cè)面其他方向。同時，橫梁的一端可以在軸向自由轉(zhuǎn)動（水平方向）。簡單橫梁的支撐可能會受到向上或者向下的反作用力。3 在圖1.2（b）中，橫梁的一端是固定的，另一端是自由懸空的，這就是所謂的懸空梁。橫梁被固定的一端既不能旋轉(zhuǎn)也不能移動，但是另一端則是可以旋轉(zhuǎn)和移動的。圖中的第三個例子展示了含有伸出部分的橫梁。這個橫梁在圖a和b中被簡單支撐同時在c中有了自由端。4 在橫梁上的負載可以是垂直的，如圖1.2（a）和（c）中的力p1和p2，

19、或者是平均分散的負載，如圖1.2（b）中的負載q. 平均分散的負載是以他們的密度分布來區(qū)分，這是以在軸線方向上單位長度上得力大小。對于同意的分散負載，如圖1.2（b）中所示，其強度為常數(shù)；一個變化的負載，在另一方面，是在軸線長度方向上功能強度發(fā)生變化的。5 在圖1.2所示的橫梁均為靜止確定的，因為它們所有的受力后作用效果均可從靜態(tài)平衡方程中得到確定。例如，在支撐負載p1的簡單橫梁所示的情況中，所有的作用效果都是在垂直方向的，并且其大小也可以通過總結(jié)受力完成瞬間來確定；而且，我們可知： 同理我們可以找出確定圖1.2（c）中所示的有外伸端的橫梁作用效果。6 對于懸梁圖1.2(b)，如圖中所示，應用

20、負載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。從垂直方向上力的總結(jié)可得，并且通過對質(zhì)點a的瞬間的總結(jié)可得ma的工作效果如圖所示，是延時鐘方向。7 先前圖示的例子說明靜態(tài)確定橫梁的作用效果可以從方程中計算出來。靜態(tài)非確定橫梁的作用效果的確定需要考慮橫梁的形變效果，這種研究將在以后的課程學習中討論。8 圖1.2中所示的理想化支撐條件只是偶爾在實際中才會遇到。舉例說明如，在橋的大跨度橫梁兩端有時候也被建成用鉸和滾動支撐。當然，在短一點的橫梁上，經(jīng)常會有對支撐的水平移動的制約力。在大部分情況下，這種制約對橫梁作用效果有很小的影響，是可以被忽略的。當然，如果橫梁是非常容易彎曲的，并且假如兩

21、端水平的制約力是非常有效果的，那就有必要考慮它們共同的作用效果。 例題 找出圖1.3（a）中所示的簡單橫梁受力下的反作用力，忽略橫梁自身的重量。 解答 橫梁所受到的力已經(jīng)在圖示中給出。支撐力的性質(zhì)接下來就會測出，并且這些部分中的未知組成部分被大膽的體現(xiàn)在圖示中。有未知反作用力組成部分和所有已提供力的橫梁被重新展示在圖1.3（b）中，來刻意的強調(diào)構(gòu)建這個自由體圖示的重要步驟。在a中，由于一端是被別住的，或許存在兩個未知的反作用力。由于b端是在滾動作用上，所以其反作用力只可能在垂直作用力上。所有力的作用點都被認真標記出。當一自由體的受力圖被完成之后，需要有靜態(tài)方程來求解，rax=0，2000+10

22、0（10）+160（15）rb=0，rb=+2700lb,ray(20)+2000100（10）160（5）=0，ray=10lb驗證：，10100160+270=0 應注意包含了三個非獨立靜態(tài)方程中的一個，所以只有兩個附加的反應力組成部分可能從方程中被求出。假如有更多的反應力的組成部分或是瞬間存在于支撐中，問題成了靜態(tài)不可解性的。 注意在點c中心的力集中的瞬間只是出現(xiàn)在所有瞬間總和的表現(xiàn)中。rb正向標記說明rb已經(jīng)在圖1.3（b）中被正確的假定。相反的是ray的情況，并且a點的垂直反作用力是向下的，注意假如計算如上所示，那么計算過程中的驗證是有效的。2.share fore and bend

23、ing moment in beamslet us now consider, as an example, 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 th

24、e beam (that is, the right-hand part) upon the left 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 lo

25、ad p. it convenient to resolve the resultant into 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 bend

26、ing moment acting at a cross section of a beam are 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 diagra

27、m shown in the second part of the figure. from summations 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 secti

28、on mn,。thus, through the use of a free-body diagram 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 stress

29、es associated with bending moment m and the shear 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 consid

30、ered, we will find that the stress resultants have 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 dir

31、ection with respect to the material against which 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 ac

32、ts (tension), a positive shear force acts clockwise about 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 load, a concentrated force p and a couple mo, acting as shown in fig.1.7（a）. find the shear f

33、orce and bending moment in the beam at cross sections 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.solution the first step in the analysis of this beam is to find the reaction ra and rb .taking moment a

34、bout ends and a and b give two equations of equilibrium, from which we find next, 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 fi

35、g.1.7(b).the force p and the reaction ra appear in this 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

36、of force in the vertical direction gives v=which shows 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 t

37、his equation, we see that the bending moment m may be either 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 a

38、nd the former one is that the couple mo now acts on the 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 th

39、e cut section is shifted from left to right of the couple mo, but the bending moment increases algebraically by an amount equal to mo.梁的剪力和彎矩讓我們來研究，一個例子，自由端上作用有傾斜負載p的懸臂梁，如圖（1.5a）所示。如果我們在截面mn處切斷并將梁的左端隔離作為一個自由體，如圖（1.5b）所示，我們看到，被隔離出來的那部分橫梁（即右半部分）必須和左半部分一樣是處于平衡狀態(tài)。在截面mn上的應力分布，在我們現(xiàn)階段的學習中是不可知道的，但是我們可以知道應力的

40、合力一定與負載p平衡。這樣我們就很方便去解決由此而產(chǎn)生的一個軸向力n作用于截面和通過軸心的橫截面，平行于截面的剪力v和作用在梁的平面上的彎矩m。作用在梁截面的軸向力，剪力和彎矩都是可知道的內(nèi)力。對于一個靜態(tài)梁，內(nèi)力可以由平衡方程來決定。因此，對于如圖（1.5）的懸臂梁，我們可以寫出三個關(guān)于自由體圖解的靜態(tài)方程，如圖第二部分所示。水平和垂直方向的合力，我們可以發(fā)現(xiàn)，分別地為， n=pcos v=psin通過mn截面形心的軸的彎矩和為m=pxsinx為自由端到截面的距離。因此，通過運用自由端的圖解和靜態(tài)平衡方程，我們可以很容易地計算出內(nèi)力。由梁上軸向力n單獨作用引起的應力在第二單元的章節(jié)已經(jīng)討論過

41、?，F(xiàn)在我們需如何獲得與應力相關(guān)的彎矩m和剪力v。我們假定內(nèi)力n、v和m的方向是正向的，它們的作用方向如圖1.5b所示。這一規(guī)定是有用的，但是，只能是在我們在討論梁左端的平衡時。如果討論梁的右端，我們將會發(fā)現(xiàn)內(nèi)力有相同的作用效果，但是方向是相反的（看圖1.5b）。因此，我們必須承認內(nèi)力的代數(shù)和是不依賴于它的空間方向的，如向左或者是向右，而是依賴于作用在材料上的反方向。為了說明事實，對n、v、m的規(guī)定在圖1.6有復述，即為力作用于橫梁上產(chǎn)生的效果。我們發(fā)現(xiàn)正面的的軸力反作用于其表面（拉力），正向剪力順時針作用于表面，以及正面的彎矩是作用在壓縮梁上部的正向彎矩之一。例題 一個簡單梁ab有著兩個負載，

42、一個集中負荷p和一個力偶mo，作用如圖1.7a所示。找出梁上截面處的的剪力和彎矩，如下：（a）梁中間到左端的一小段距離(b)梁中點到右端的一小段距離。解答 這個梁的分析的第一步是找出力ra和rb。取a、b端可列出兩個平衡方程，我們發(fā)現(xiàn)， 接下來，將梁在截面出剪斷并僅取中點的左端，將梁的一半作為自由體做圖解。在這個例子中，我們選擇梁的左半邊為研究對象，其相應的圖解如1.7b所示。負荷p和力ra出現(xiàn)在這個圖解上，同樣也有未知的剪力v和彎矩m，它們都是為正方向的，力偶則不出現(xiàn)在這個圖上，因為梁被切斷點的左端沒有mo作用。在垂直方向的合力為 v=這說明剪力是負的；今后，假定其作用在反方向上，如圖1.7

43、b。取通過分截面對軸的矩，如圖1.7b所示，則有 m=在這個公式中，m依賴于相關(guān)的代數(shù)和，我們可以看到m可能是負的或者是正的。為了獲得截面的內(nèi)力，僅取梁的右端為研究對象，我們在梁的截面處切斷，然后重新畫自由體圖解1.7c。這個圖解跟前面那個的不同點僅僅是現(xiàn)在力偶mo作用于切斷面的左端上。重新求垂直方向上的合力，同時求切斷面對軸的彎矩，我們可以得到 v= m=我們從這些結(jié)果可以看到，當切斷面由力偶mo左端轉(zhuǎn)換到右端，剪力都不會改變，但是彎矩的代數(shù)和增量等于mo。3.theories of strength1. principal stresses the state of the stress

44、at a point in a structural member under a complex system of loading 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-dimensiona

45、l stress system, fig.1.11, the principal stresses at any pint are related 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 princip

46、al stresses: maximum shear stress,compressive stresses are conventionally 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 diam

47、eter: thin-wall 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.1.12. if the wall is thin, the radial stress 3 will be small and can be neglected in compar

48、ison with the other stresses , and the longitudinal and circumferential stresses1 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

49、 in the chemical and allied industries are classified as thin-walled vessels. 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 equa

50、tions can also be used to select the size of a member if the members strength 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 materi

51、al will rupture when subjected to a purely axial load. this property is determined 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. t

52、he ultimate stress or strength is the 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. in the most engineer

53、ing structures it is desirable that the structure not fail. thus design is based 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

54、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 and for the actual stress in material. in a design: sa this so-called factor of safety co

55、vers a multitude of sins. it includes such factors as the uncertainty of the load, 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

56、 lower the factor of safety necessary.4. theories of failurethe 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 tensile test, but for components subjected to combined stresses (normal and shear stress) the position is not so simple, and several theories of failure have been proposed. the three theories most commonly used are described below: maximum principal stress theory: which postulat