过程装备与控制系统工程专业英语 - 图文

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过程装备与控制工程

专业英语

学院：化学化工学院

1.Static Analysis of Beams

⑴ 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 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 consider a more general kind of bending in which the beam may have an unsymmetrical cross section.

⑵ 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

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a simple beam may sustain vertical reactions acting either upward or downward .

⑶ 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.

⑷ 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 along the axis of the beam.

⑸ 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 find RA?P1(L?a)P1L RB? LLThe reactions for the beam with an overhang [Fig.1.2 (c)]can be found the same manner.

⑹ For the cantilever beam[Fig.1.2(b)], the action of the applied load q is

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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

RA?qb，

And ,from a summation of moments about point A, we find

bMA?qb(a?)，

2The reactive moment MA acts counterclockwise as shown in the figure. ⑺ 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.

⑻ 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.

⑼ 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.

⑽ Solution The loading of the beam is already given in diagrammatic form. The

nature of the supports is examined next and the unknow components of

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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.

?Fx?0，RAx=0

?0?，2000+100（10）+160（15）—RB=0，RB=+2700lb↑ ?0?,RAY(20)+2000—100（10）—160（5）=0，RAY=—10lb↓

?M?MABCheck：⑾ Note that

?FX?0??，—10—100—160+270=0 ?Fx?0uses 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.

⑿ 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.

横梁的静态分析

⑴ 一条绕其轴水平放置的棒就是所谓的横梁，本章节我们将研究最简单的横梁模型形式，

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如图1.2所示。在每个例子中，假设横梁有平整的外形，并且都与其轴线平衡。并且，交叉的部分都有着外形相垂直的关系，同时假设那些负载均作用在平整的外形面上，所以弯曲将会发生在平整的横梁上。稍后我们将研究一个非常普遍常见的并且有垂直于非平整部分的弯曲。

⑵ 在图1.2（a）的横梁中，横梁的一端由铰链支座支撑着，另外一端由滚动支座支撑，

这就是所谓的简单的支撑横梁，或者简单横梁。此简单横梁的特征为两端在弯曲过程可以自由转动，但是不能被转动到侧面其他方向。同时，横梁的一端可以在轴向自由转动（水平方向）。简单横梁的支撑可能会受到向上或者向下的反作用力。

⑶ 在图1.2（b）中，横梁的一端是固定的，另一端是自由悬空的，这就是所谓的悬空梁。

横梁被固定的一端既不能旋转也不能移动，但是另一端则是可以旋转和移动的。图中的第三个例子展示了含有伸出部分的横梁。这个横梁在图A和B中被简单支撑同时在C中有了自由端。

⑷ 在横梁上的负载可以是垂直的，如图1.2（a）和（c）中的力P1和P2，或者是平均

分散的负载，如图1.2（b）中的负载q. 平均分散的负载是以他们的密度分布来区分，这是以在轴线方向上单位长度上得力大小。对于同意的分散负载，如图1.2（b）中所示，其强度为常数；一个变化的负载，在另一方面，是在轴线长度方向上功能强度发生变化的。

⑸ 在图1.2所示的横梁均为静止确定的，因为它们所有的受力后作用效果均可从静态平衡