过程装备与控制工程课后材料专业英语翻译.doc

装控091 专业英语翻译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 symmetry 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 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 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 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.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. 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 determinate 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 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 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 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.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 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 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.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 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 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, thus 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 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 caculations are made as shown.Reading materia l 一条受力作用沿着横向坐标的轴类称作横梁。在这个部分，我们只考虑一些最简单的横梁，例如那些展现的图1.2中的。假设每一个横梁都有水平面对称，和自身的线平行对称，因此，横梁的横截面又垂直的对称性。而且，假设负载沿水平方向均匀作用在横梁上，弯曲将会发生在水平面上。然后我们将要讨论那些更常见的不对称横截面横梁的弯曲 1.2 a 中的横梁，由一端固定另一端滚动组成的横梁，称为简单支撑横梁，或简单横梁，简单横梁主要的特征是两个末端在弯曲时都可以自由转动，但是不能转换。简单横梁可以支撑向上或向下的反力。图1.2 b 中的一端固定，一端自由运动的横梁，称为悬臂支撑横梁，固定端悬臂支撑不能旋转和转化，但是自由端可以。图表例子3中显示的是外伸横梁，由固定端A B和自由端C支撑。 负载加载在横杆上的力有可能是集中的载荷。就像图1.2a 1.2c 的P1 P2 ，如加载在例子1.2b中，分布载荷的特点在于他们的集中度已沿着横梁轴向单位距离表示力的单位，为了在例子1.2b中图解一致性的分布载荷，且其变动载荷的强度是连续的，另一方面，强度大小的变化和横梁的轴向距离变化有关。 例1.2中的横梁是静定的，因为所有的力都可以通过平衡方程求出。例如1.2a中的简单横梁的负荷，两个反作用力都是垂直的，其大小也可以通过总结两边受力瞬间来确定。因此，我们发现 Ra = P1(L-a)/L Rb=P1a/L 外伸横梁上的反作用力也可以通过相同的方式得到对于1.2b的悬臂横梁，施加载荷使垂直力RA和力偶MA在固定端相等。如图所示，从合力方向垂直，我们可以总结 Ra =qb 还有从A点得合力来看，我们发现MA=qb(a+b/2) 反作用力偶如图逆时针方向旋转。之前的例子说明了静定横梁的作用效果可以通过方程计算静不定的横梁作用效果需要考虑到横梁的形变效果，此处这种研究方法教材以后会讨论。图1.2中的理想化支撑条件只是偶尔在练习中遇到。举例，桥的大跨度横梁有时也会建造成铰链和滚动支撑。然而，在短一点的横梁，经常会有对支撑水平移动的制约力。大部分情况下，这种制约对横梁的效果很小，可以被忽略。当然，如果横梁非常容易弯曲，并且假如两端水平的制约力是非常有效果，那就又必要考虑他们共同作用的效果例子找出图1.3 a 中的简单横梁受力下的反作用力，忽略横梁自身的重量横梁受力已经在图中给出。测试支撑力的性质和横梁的未知部件明显地在图中展示。有未知的反作用力和所有提供的力重新画在1.3b ，来可以强调构建这个自由体的重要步骤，在A中，由于一端是固定住，可能存在两个未知的反作用力。由于B端滚动，其他在只可能在垂直作用力上。所有的力量被认真标记处。当这个自由体的受力图表完成后，这个问题需要用静态方程求解。Fx=0 Rax=0MA=0+,2000+100(10)+160(15)-Ra(20)=0,RB=+2700lbMb=0+, Ray(20)+200-100(10)-160(5)=0 Ray=-101b验证Fy=0， -10-100-160+270=0注意 Fx=0 用到了 3个非静态方程中的一个，所以只有两个附加的反应力组成部分可能由方程中可以求出。但是如果有更多的反应力存在支撑结构中，问题成了静不定。 注意点C的中心的力集中瞬间只是出现在所有瞬间总和的表现。Rb的正向标记说明在1.3b中做出了正确的假设。Ra的情况相反，并且A点得垂直反作用力是向下的。注意假如计算如上那么计算过程中的验证是有效的。UNIT 2 Reading material 2Shear Force and Bending 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 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. 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 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 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 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 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 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.1.5(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 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 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 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). Find 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 . Taking 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 diagram 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 is 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 relative 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 difference 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 force 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.现在让我们考虑，举例，一个倾斜的载荷P作用于一个悬臂杆的自由端。假如我们穿过杆件的横截面 MN，使得左手边部分脱离杆件成为自由体，我们发现右边部分必须与左边部分保持力的平衡。在我们的分析学习中，我们并不知道压应力在mn面的分配情况。但我们的确知道这个压应力是与载荷P相平衡的。们可以简单地把其分解为一个通过轴心的轴向力N，一个和截面平行的剪应力，和一个作用在轴上得弯矩M。应力的必然结果是在截面产生了轴向力，剪应力和弯矩。对于一个静定的杆件，可由这三个要素确定其平衡方程。因此图中的悬臂梁，我们可以得到三个平衡方程，从受力图分析，水平和垂直两个方向上，我们可以得到以下两个方程，其依次为：N=Pcos V=Psin从通过质心截面我们总结M=Pxsin，X表示自由端到MN的距离。可见，通过对自由体进行受力分析以及解析方程，可以轻易地求取轴向力，剪切力和弯矩。对于轴应变力N是轴向应力单独作用的结果，现在我们将会发现压力和弯矩和剪应力V的关系 我们把N ,V,M但他们如图所示方向都设为正方向。当我们考虑梁的左边部分时，这样的分析方法很方便。对右边部分的考虑，结果是类似的，只是方向相反而已，如图see Fig.1.5(c).所示。因此，我们必须要注意到这些应力的代数关系并不依赖于它们的空间方向，而是取决于材料对它的作用。为了证明这一点，我们可以对N、V和M进行微元分析，如图Fig.1.6所示。我们规定垂直截面向外的方向为轴力的正方向，顺时针方向为剪应力的正方向，而使得梁的上部材料被压的弯矩方向为正方向。例子简支梁AB受两个载荷，一个集中力P和一对力偶Mo，如Fig.1.7（a）。剪力和弯矩在梁截面的分布的特征是左右边都与中心面相距很小距离。解决方法第一步，先分析Ra和Rb，分别对A、B两点列力矩平衡方程，有Ra=3P/4-Mo/L Bb=P/4+Mo/L下一步，用一个假想的平面将梁分开，有以下两图。选左边来研究，如图 Fig.1.7(b)，在我们的研究体系中存在力P和Ra，以及未知剪力V和弯矩M，两个均为正方向，因为我们只是取左边部分考虑，所以在这分析中不考虑Mo，故在垂直方向的合力为V=Ra-P=-P/4-Mo/L剪应力为负值，说明它与假定方向相反。对被截面形心列力矩式，有M=RaL/2-PL/4=PL/8-Mo/2鉴于其中的相对性，方程中所求出的力矩有可能是正值，也有可能是负值。为得到右半部分所受应力情况，同样是将其从中间分开，有图Fig.1.7(c). 此图跟前一图的区别就是力矩Mo已经考虑在梁上。再次在垂直方向列力的平衡方程以及对截面形心列力矩平衡方程，有V=-P/4-Mo/L M=PL/8+Mo/2从结果分析，有以下结论：力矩Mo在梁上左右移动时，剪切力并没有改变，但弯矩和Mo成线性比例关系。unit 3 力学理论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 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 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 principalstresses:Maximum shear stress,Compressive stresses are conventionally taken as 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-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 comparison 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 in the chemical and allied industries are classified as thin-walled vessels. Thick-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 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 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. The 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 engineering 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 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: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 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 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 strength of the material divided by a suitable factor of safety. Maximum shear stress theory: which postulates that failure will occur in a complex stress system when the maximum shear stresses reaches the value of the shear stress at failure in simple tension.For a system of combined stresses there are three shear stresses maxima:， （1.10）In the tensile test, 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 maximum value