2019年过程装备与控制工程专业英语翻译阅读材料1.doc

Reading material 1

Static Analysis of Beams

Note that ∑=0x F uses up one of the three independent equations of statics, thus only two additional reaction components may be determinated from statics. If more unknow reaction components or moments exist at the support, the problem becomes statically indeterminate.

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.

The beam in Fig.1.2（a), 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 they cannot translate in 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 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 , such as the the load q in Fig.1.2(b),。Distributed loads are characterized by their intensity,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.

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 P 1 [Fig.1.2(a)], both reactions are vertical, and their magnitudes can be found by summing moments about the ends; thus, we find

L

a L P R A )(1-= L L P R B 1= The reactions for the beam with an overhang [Fig.1.2 (c)]can be found the same manner For the cantilever beam[Fig.1.2(b)], the action of the applied load q is equilibrated by a vertical force R A and a couple M A acting at the fixed support, as shown in the figure. From a summation of forces in vertical direction , we include that

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

The reactive moment M A acts counterclockwise as shown in the figure

The preceding examples illustrate how the reactions(forces and moments) of statically determinate beams may be calculated by statics.The determination of the reactions for statically

indeterminate 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.

Solution

The loading of the beam is already given in diagrammatic form. The nature of the supports is examined next and the unknown 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 the end is pinned. The reaction at B can only act in a vertical direction since the end is on a 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 obtain the solution.

Note that the concentrated moment applied at C enters only the expressions for summation moments. The positive sign of R B indicates that the direction of R B has been correctly assumed in Fig.1.3(b). The inverse is the case of R AY ,and the vertical reaction at a is downward. Noted that a check on the arithmetical work is available if the calculations are made as shown.

阅读材料1

梁的静力分析

在其轴线上受到横向力的杆我们称之为梁。在此，我们仅一些最简单的类型，如图1.2所示情形作分析。在每一例中，我们假定有一平面平行于物体表面。从而，横梁的交叉部分就有一条垂直的轴线。同时假设有力作用于平行面上，导致该面发生弯曲。而后，我们将这种情形推广至一种更普遍的即横梁没有对称交叉部分的情况。

如图1.2（a）所示的一端受销钉支撑、另一端受滚动支座支撑的杆我们称之为“简力支撑杆”或“简支杆”。简支杆的基本特征是其两端都可自由旋转，但不可横向移动。或者杆的一端可以在轴线方向上（即水平方向）自由移动。简支杆的支撑端可以维向上或向下的垂直反力。

如图1.2(b)中一端固定，另一端自由的杆称为“悬梁”。其一端既不可转动也不可移动，另一端则完全相反。第三个图所示为一伸出杆。其在A、B端受支撑，C端自由。

梁的负载有可能是集中力，比如图1.2（a）1.2（c）中的P1和P2，或是分散力，比如图1.2（c）中的q。

分散力往往通过其密度来表述，单位是在梁的轴向方向上每单位长度的受力大小。均匀规律分布的分散负载，如图1.2（b）所示，其密度固定不变；而对于变化的负载，其密度作为梁的轴向上的函数而随时改变。

图1.2中的梁均可静态确定因为它们所受的反作用力均可根据静态平衡方程确定。例

如，在如图1.2（a ）的简支梁的负载中，其反作用力均为垂直的，其大小均可通过统计两端的力矩得出。由此我们得到

R A = P 1(L-a)L R B =P 1a L

如图1.2（c ）中所示带有伸出部分的梁的反作用力可以用同样的方法得出。

图1.2（c ）中的悬梁，应用负载在一竖直力R A 和一力偶M A 的作用下使梁保持平衡。对竖直方向上的力作统计，我们得出

R A =qb

同样，对点A 所受力矩做统计，得出

M A =qb(a+b 2

) 反力偶M A 逆时针方向作用。

前面的例子表述的是怎样通过静态分析将静态确定梁的反作用力（力和力偶）计算出来。而静态不确定梁的判定需要考虑梁的弯曲，因此物体本身被置于次要地位。

如图1.2中的理想化支持力条件在实际中只会偶尔遇到。例如桥梁的大跨度梁有时需要安装销钉和滚动支座。然而，在一些稍短的梁中常会产生一些约束支持水平力偶的力。大多数情况下这些力产生的效果微乎其微，可以忽略。但如果在梁很容易弯曲或其两端的水平约束力很坚固，即有必要考虑其影响。

【例】

求出如图1.3（a ）简支杆支撑端的反作用力。忽略杆重。

【解】

梁的负载已经在受力简图中给出。自然支持力已在旁边给出分析，未知力的部分也已清晰标出。该梁以及其未知的反作用力和应力均在一独立简图1.3（b ）中重新标出以起强调该步骤的作用。在A 点存在两未知反作用力，因为该处有销钉固定。B 点的反作用力只在竖直方向上起作用，因为该处置于滚动支座上。所有的受力点均仔细标出。梁的受力简图画好后，既要列出静态方程以得出答案。

∑Fx=0,R Ax =0

注意到∑Fx=0使用了三个静态独立方程中的两个，因此，从该静态方程中仅可以确定两个另外的反作用力。如果还有更多的反作用力或力矩，该问题则变为静不定问题。

注意到C 点集中应力仅在力矩统计方程中出现。R B 的正号表示其方向正是图1.3（b ）中标明的方向。R Ay 的情况则相反，即A 点竖直方向受向下的作用力。注意数学运算上的检查。