过程装备与控制工程专业英语翻译

1、过程装备与控制工程专业英语翻译 专业：过程装备与控制工程姓名： 学号： reading material 1 static analysis of beams a bar that is subjected to forces acting transverse to its axis is called a beamin this section we will consider only a few of the simplest types of beams，such as those shown in fig12in every instance it is assumed that

2、the beam has a plane of symmetry that is parallel to the plane of the figure itselfthus，the cross section of the beam has a vertical axis of symmetryalso，it is assumed that the applied loads act in the plane of symmetry，and hence bending of the beam occurs in that planelater we will consider a more

3、general kind of bending in which the beam may have an unsymmetrical cross section (a)a simple supported beam (b) a cantilever beam (c) a beam with an overhang fig12 types of beams the beam in fig12(a)，with a pin support at one end and a roller support at the other，is called a simply supported beam，o

4、r a simple beamthe essential feature of a simple beam is that both ands of the beam 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 sustai

5、n vertical reactions acting either upward or downward. the beam in fig. 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

6、figure shows a beam with an overhang. this beam is simply supported at a and 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),or distributed loads, such as the load q in fig. 1.2(b). distributed loads are characterized by their intensit

7、y, which is expressed in units of force per unit distance 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 along the axis of the beam.

8、 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 plfig. 1.2(a), both reactions are vertical, and their magnitudes can be found by summing moment

9、s about the ends; thus, we find ra=p1 (l-a)/ l rb =p1a/ l the reactions for the beam with an overhang fig. 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 suppo

10、rt, as shown in the figure. from a summation of forces in the vertical direction, we conclude that ra =q b and, from a summation of moments about point a, we find ma =q b(a+b/2) the reactive moment ma acts counterclockwise as shown in the figure. the preceding examples illustrate how the reactions(f

11、orces and moments)of statically determinate beams may be calculated by staticsthe determination of the reactions for 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 fig12 are encoun

12、tered only occasionally in practice. as an example, longspan beams in bridges sometimes are constructed with pin and roller support at the ends. however, in beams of shorter span, there is usually some restraint against horizontal movement of the supportsunder most conditions this restraint has litt

13、le effect on the action of the beam and can be neglectedhowever，if the beam is very flexible, and if the horizontal 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 fig13(a) neglect t

14、he weight of the beam solution the loading of the beam is already given in diagrammatic formthe nature of the supports is examined next and the unknown components of these reactions are boldly indicated on the diagramthe beam，with the unknown reaction components and all the applied forces，is redrawn

15、 in fig13(b)to deliberately emphasize this important step in constructing a freebody diagramat a，two unknown reaction components may exist，since the end is pinnedthe reaction at b can only act in a vertical direction since the end is on a rollerthe points of application of all forces are carefully n

16、otedafter a free body diagram of the beam is made，tile equations of statics are applied to obtain the solution fig1，3 a simple beam fx=0， rax=0 ma=o+，2000+100(10)+160(15)-rb(20)=0，rb=+2700 lb mb=o+，ray(20)+2000-100(10)-160(5)=0，ray=-10lb check：fy=0+，-10-100-160+270=0 note that fx=0 uses up one of th

17、e three independent equations of staticsthus only two additional reaction components may be determined from staticsif more unknown reaction components or moments exist at the support，the problem becomes statically indeterminate note that the concentrated moment applied at c enters only into the expr

18、essions for the summation of momentsthe positive sign of rb indicates that the direction of rb has been correctly assumed in fig13(b)the inverse is the case of ray, and the vertical reaction at a is downwardnote that a check on the arithmetical work is available if the calculations are made as shown

19、 (selected from stephen p. timoshenko and james m. gere,mechanics of materials* van nostrand reinhold company ltd. ,1978. * selected from egor p. popov, introduction to mechanics of solids*prentice-hall inc. ,1968. ) 材料1 横梁的静力分析 一条受到由截面向轴心的力的棒子称为横梁。在这文章中我们将讨论几种最简单的受力图，如图1.2所示。每种情况下，我们假设横梁水平对称即与其平整的外

20、形对称。因此,横梁截面梁垂直对称轴。同时假定施加载荷于对称平面，同时弯曲也发生在那个平面。之后，我 们考虑一种更普遍的弯曲，可能有非对称截面。 （a）简单的支撑横梁 （b）悬臂梁 （c）外伸梁 图1.2 横梁的种类 在图1.2（a）中杆一端由固定支座支撑另一端由一个滚动支座支撑，被称为简支梁或者单体梁。简支的一个基本的特征是在弯曲的时候杆的两端可以自由旋转，但是不能横向移动。杆一端能够在轴向自由移动（意思是水平方向）。简支梁的支撑可能垂直向上或者向下。如图1.2（b）杆一端固定，另一端自由的称作悬臂梁。在固定的一端既不能旋转也不能移动，但另一端两者均可。图中第三个例子展示的是带有悬垂部分的外伸

21、梁。梁在ab处受到支撑，有一个自由端c。杆上的载荷可能集中在一点，例如图1.2（a）的p1和1.2（c）的p2，或者分散分布，例如,图1.2（b）的q. 具有分布式荷载强度,主要表现为一个单位距离力沿杆的主轴。对于均匀呢分布的载荷，例如1.2（b）中，其强度是不变的；非均匀分布的载荷其强度是随沿其杆轴线方向距离的一个函数。 图1.2的横梁是静定的，因为所以的反应都可以从静力平衡方程中得出。例如，图1.2（a）中简支梁受的力p1，两个方向的受力都是垂直的，并且其大小也可以通过总结受力完成瞬间得出，因此，我们发现ra=p1（l-a）/l rb=ap1/l 图1.2（c）的外伸梁的受力也可以用同种方法获得。 对于图1.2（b）的悬臂梁，所受应用载荷q与一个垂直力和固定端的力偶平行，如图所示。从以上垂直方向的合力，我们总结得出ra = qb 。从a点的合力，我们发现ma=qb(a+b/2)，作用力力偶是逆时针的如图所示。 前面的例子说明了静定力能够由静力学方程求出来。静力不确定横梁的力需要考虑梁的变形，这个在以后会做讨论。 图1.2中的理想化条件在实际中只是偶尔遇到。例如，大桥中的长距离横梁两端有时候 需要铰链和滚动结