過程裝備與控制工程課后材料專業(yè)英語翻譯

1、 裝控091 專業(yè)英語翻譯unit 1Static Analysis of BeamsA bar that is subjected to forces acting trasverse to its 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 Flag.1.2. In every instance it is assumed that the beam has a plane of symmet

2、ry that is parallel to the plane of the figure itself. Thus the cross section 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 cross section.The beam in Fig.1.2, with a pin support at one end and a roller support at t

3、he 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 the cannot translate in lateral direction. Also ,one end of the beam can move freely in the axial direction (that is, horizontal). T

4、he 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 beam. At the fixed support the beam can neither rotate nor translate, while at the free end it

5、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.1.2(a) and (c), or distributed loads loads, such as the the load q in Fig.1.2(b), the intesity

6、. 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 other hand, is one in which the intensity varies as a function of distance alongthe axis of the beam.The beams shown in Fig.1.2 are statically deter

7、minate because all their reactions can be determined from equations of static equilibrium. For instance ,in the case of the simple beam supporting the load P1 Fig.1.2(a), both reactions are vertical, and tehir magnitudes can be found by summing moments about the ends; thus, we findThe reactions for

8、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 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 incl

9、ude 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 illustrate how the reactions(forces and moments) of statically determinate beams requires a considerition of the bending of the beams , and henc

10、e 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 constructionn with pin and roller supports at the ends. However, in beams of shorter span ,there is usually some r

11、estraint 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, if the beam is very flexible, and if the horizonal restraints at the ends are very rigid , it may be necessary to consider their effect

12、s.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.SolutionThe loading of the beam is already given in diagrammatic form. The nature of the supports isexamined next and the unknow components of reactions are boldly indicated

13、on the diagram. The beam , 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 roller. The points of application of all forces

14、 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=+2700lb,RAY(20)+2000100（10）160（5）=0，RAY=10lbCheck：，10100160+270=0Note that uses up one of the three independent equations of statics, t

15、hus only two additional reaction compones may be determinated from statics. If more unknow reaction components or moment exist at the support, the problem becomes statically indeterminate.Note that the concentrated moment applied at C enters only the expressions for summation moments. The positive s

16、ign 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 caculations are made as shown.Reading materia l 一條受力作用沿著橫向坐標的軸類稱作橫梁。在這個部分

17、，我們只考慮一些最簡單的橫梁，例如那些展現(xiàn)的圖1.2中的。假設每一個橫梁都有水平面對稱，和自身的線平行對稱，因此，橫梁的橫截面又垂直的對稱性。而且，假設負載沿水平方向均勻作用在橫梁上，彎曲將會發(fā)生在水平面上。然后我們將要討論那些更常見的不對稱橫截面橫梁的彎曲 1.2 a 中的橫梁，由一端固定另一端滾動組成的橫梁，稱為簡單支撐橫梁，或簡單橫梁，簡單橫梁主要的特征是兩個末端在彎曲時都可以自由轉動，但是不能轉換。簡單橫梁可以支撐向上或向下的反力。圖1.2 b 中的一端固定，一端自由運動的橫梁，稱為懸臂支撐橫梁，固定端懸臂支撐不能旋轉和轉化，但是自由端可以。圖表例子3中顯示的是外伸橫梁，由固定端A B

18、和自由端C支撐。 負載加載在橫桿上的力有可能是集中的載荷。就像圖1.2a 1.2c 的P1 P2 ，如加載在例子1.2b中，分布載荷的特點在于他們的集中度已沿著橫梁軸向單位距離表示力的單位，為了在例子1.2b中圖解一致性的分布載荷，且其變動載荷的強度是連續(xù)的，另一方面，強度大小的變化和橫梁的軸向距離變化有關。 例1.2中的橫梁是靜定的，因為所有的力都可以通過平衡方程求出。例如1.2a中的簡單橫梁的負荷，兩個反作用力都是垂直的，其大小也可以通過總結兩邊受力瞬間來確定。因此，我們發(fā)現(xiàn) Ra = P1(L-a)/L Rb=P1a/L 外伸橫梁上的反作用力也可以通過相同的方式得到對于1.2b的懸臂橫梁

19、，施加載荷使垂直力RA和力偶MA在固定端相等。如圖所示，從合力方向垂直，我們可以總結 Ra =qb 還有從A點得合力來看，我們發(fā)現(xiàn)MA=qb(a+b/2) 反作用力偶如圖逆時針方向旋轉。之前的例子說明了靜定橫梁的作用效果可以通過方程計算靜不定的橫梁作用效果需要考慮到橫梁的形變效果，此處這種研究方法教材以后會討論。圖1.2中的理想化支撐條件只是偶爾在練習中遇到。舉例，橋的大跨度橫梁有時也會建造成鉸鏈和滾動支撐。然而，在短一點的橫梁，經(jīng)常會有對支撐水平移動的制約力。大部分情況下，這種制約對橫梁的效果很小，可以被忽略。當然，如果橫梁非常容易彎曲，并且假如兩端水平的制約力是非常有效果，那就又必要考慮他

20、們共同作用的效果例子找出圖1.3 a 中的簡單橫梁受力下的反作用力，忽略橫梁自身的重量橫梁受力已經(jīng)在圖中給出。測試支撐力的性質和橫梁的未知部件明顯地在圖中展示。有未知的反作用力和所有提供的力重新畫在1.3b ，來可以強調構建這個自由體的重要步驟，在A中，由于一端是固定住，可能存在兩個未知的反作用力。由于B端滾動，其他在只可能在垂直作用力上。所有的力量被認真標記處。當這個自由體的受力圖表完成后，這個問題需要用靜態(tài)方程求解。Fx=0 Rax=0MA=0+,2000+100(10)+160(15)-Ra(20)=0,RB=+2700lbMb=0+, Ray(20)+200-100(10)-160(5

21、)=0 Ray=-101b驗證Fy=0， -10-100-160+270=0注意 Fx=0 用到了 3個非靜態(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由方程中可以求出。但是如果有更多的反應力存在支撐結構中，問題成了靜不定。 注意點C的中心的力集中瞬間只是出現(xiàn)在所有瞬間總和的表現(xiàn)。Rb的正向標記說明在1.3b中做出了正確的假設。Ra的情況相反，并且A點得垂直反作用力是向下的。注意假如計算如上那么計算過程中的驗證是有效的。UNIT 2 Reading material 2Shear Force and Bending Moment in BeamsLet us now consider,

22、 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 the beam (that is, the right-hand part) up

23、on the left-hand 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 is convenient to resolve

24、 the resultant into an axial force N acting normal to the 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 cr

25、oss 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 diagram shown in the second pa

26、rt 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 through the centroid of cross section mn, we obtainwhere x is the distance from the free end to section mn. Thus, through the use of

27、 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 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 mo

28、ment 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.1.5(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 bea

29、m is considered, 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

30、 upon its direction 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 u

31、pon which it acts (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 loads, a concentrated force P and a couple Mo,acting as shown in Fig.1.7(a). F

32、ind the shear force 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 reactions and . Taki

33、ng moments about ends A and B gives two equations of equilibrium, from which we find Next, 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 d

34、iagram is shown in Fig.1.7(b). The force P and the reaction 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 beam is cut to the left of the point where Mo i

35、s applied. A summation of forces in the vertical direction giveswhich shows that the shear force is negative; hence, it acts in the opposite direction to 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) givesDepending upon the relati

36、ve magnitudes of the terms in this 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 d

37、ifference between this diagram and 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 forces in the vertical direction, and also taking moments about an axis through the cut section, we obtain We see from these results that the shear f

38、orce does not change when the cut section is shifted from left to right of the couple Mo, but the bending moment increases algebraically by an amount equal to Mo.現(xiàn)在讓我們考慮，舉例，一個傾斜的載荷P作用于一個懸臂桿的自由端。假如我們穿過桿件的橫截面 MN，使得左手邊部分脫離桿件成為自由體，我們發(fā)現(xiàn)右邊部分必須與左邊部分保持力的平衡。在我們的分析學習中，我們并不知道壓應力在mn面的分配情況。但我們的確知道這個壓應力是與載荷P相平衡的。

39、們可以簡單地把其分解為一個通過軸心的軸向力N，一個和截面平行的剪應力，和一個作用在軸上得彎矩M。應力的必然結果是在截面產(chǎn)生了軸向力，剪應力和彎矩。對于一個靜定的桿件，可由這三個要素確定其平衡方程。因此圖中的懸臂梁，我們可以得到三個平衡方程，從受力圖分析，水平和垂直兩個方向上，我們可以得到以下兩個方程，其依次為：N=Pcos V=Psin從通過質心截面我們總結M=Pxsin，X表示自由端到MN的距離?？梢?，通過對自由體進行受力分析以及解析方程，可以輕易地求取軸向力，剪切力和彎矩。對于軸應變力N是軸向應力單獨作用的結果，現(xiàn)在我們將會發(fā)現(xiàn)壓力和彎矩和剪應力V的關系 我們把N ,V,M但他們如圖所示方

40、向都設為正方向。當我們考慮梁的左邊部分時，這樣的分析方法很方便。對右邊部分的考慮，結果是類似的，只是方向相反而已，如圖see Fig.1.5(c).所示。因此，我們必須要注意到這些應力的代數(shù)關系并不依賴于它們的空間方向，而是取決于材料對它的作用。為了證明這一點，我們可以對N、V和M進行微元分析，如圖Fig.1.6所示。我們規(guī)定垂直截面向外的方向為軸力的正方向，順時針方向為剪應力的正方向，而使得梁的上部材料被壓的彎矩方向為正方向。例子簡支梁AB受兩個載荷，一個集中力P和一對力偶Mo，如Fig.1.7（a）。剪力和彎矩在梁截面的分布的特征是左右邊都與中心面相距很小距離。解決方法第一步，先分析Ra和

41、Rb，分別對A、B兩點列力矩平衡方程，有Ra=3P/4-Mo/L Bb=P/4+Mo/L下一步，用一個假想的平面將梁分開，有以下兩圖。選左邊來研究，如圖 Fig.1.7(b)，在我們的研究體系中存在力P和Ra，以及未知剪力V和彎矩M，兩個均為正方向，因為我們只是取左邊部分考慮，所以在這分析中不考慮Mo，故在垂直方向的合力為V=Ra-P=-P/4-Mo/L剪應力為負值，說明它與假定方向相反。對被截面形心列力矩式，有M=RaL/2-PL/4=PL/8-Mo/2鑒于其中的相對性，方程中所求出的力矩有可能是正值，也有可能是負值。為得到右半部分所受應力情況，同樣是將其從中間分開，有圖Fig.1.7(c)

42、. 此圖跟前一圖的區(qū)別就是力矩Mo已經(jīng)考慮在梁上。再次在垂直方向列力的平衡方程以及對截面形心列力矩平衡方程，有V=-P/4-Mo/L M=PL/8+Mo/2從結果分析，有以下結論：力矩Mo在梁上左右移動時，剪切力并沒有改變，但彎矩和Mo成線性比例關系。unit 3 力學理論Reading Material 3Theories of strength1. Principal stresses The state of the stress at a point in a structural member under a complex system of loading is describe

43、d 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 related to th

44、e 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,Compressive stresses are conventionally taken as

45、 negative; tensile as positive. 2. Classification of pressure vesselsFor 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 thick-walle

46、d 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 comparison with the other stresses , and the longitudinal and circumferential stresses1 a

47、nd2 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. Thick

48、-walled vessels are used for high pressures.3. Allowable stressIn 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 kno

49、wn. 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 determined f

50、rom 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 cross-sec

51、tional 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 based on

52、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 allow. T

53、he 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 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 low