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 beam．In this section we will consider only a few of the simplest types of beams，such as those shown in Fig．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．
(a)A simple supported beam (b) A cantilever beam (c) A beam with an overhang Fig．1．2 Types of beams
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 supported beam，or a simple beam．The 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 sustain 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 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 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 Pl[Fig. 1.2(a)], both reactions are vertical, and their magnitudes can be found by summing moments 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 support, 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(forces and moments)of statically determinate beams may be calculated by statics．The 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 Fig．1．2 are encountered only occasionally in practice. As an example, long—span 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 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 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 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 these reactions are boldly indicated on the diagram．The beam，with the unknown reaction components and all the applied forces，is redrawn in Fig．1．3(b)to deliberately emphasize this important step in constructing a free—body diagram．At A，two unknown 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，tile equations of statics are applied to obtain the solution． Fig．1，3 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 the three independent equations of statics．thus only two additional reaction components may be determined from statics．If 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 expressions for the summation of 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．Note that a check on the arithmetical work is available if the calculations are made as shown．
(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 需要铰链和滚动结构。当然，在短距离横梁中经常会有一些对支撑水平移动的制约力。在多数情况下，这种约束力对横梁的影响很小，可以忽略不计。但是，如果横梁的韧性横好，而两端的水平制约力非常的明显，那么就必须要考虑这个的影响了。
例子
找出图1.3(a)简单横梁受力下的反作用力。忽略横梁的重力。
解决方案
横梁的受力已经在图中给出。支撑力的性质接下来就会测出，组成部分未知的反作用力也将会在图中大胆的指出。含有未知力部分和所有应用力的横梁被重新展示在图1. 3(b)中用来着重强调构建一个自由图示的步骤。在A点，可能存在两个未知的反作用力，因为两端被约束。B点的约束反力只能在垂直方向起作用，因为B端在滑动支座上。力的作用点被仔细的标记出来。当横梁的自由结构图示制作出来，就可以利用静力学方程得出求解。 ∑Fx=0 ,RAx=0
∑MA= 0 +,2000+100(10)+160(15)-RB(20)=0,RB=+2700lb↑ ∑MB=0+,Ray(20)+2000-100(10)-160(5)=0,Ray=-10lb↓ 检查 :∑Fy=0↑+,-10-100-160+270=0
注意，∑FX= 0既是三个静力学独立方程的一个，所以只有两个额外的反作用力可以从静力学中求得。如果出现更多的反作用力部分或者更多的支持瞬间存在，问题就会变成静力学不确定。
注意，集中作用与C点的力只有一个合力瞬间的表达式。RB的正方向表明RB已经在图1.3（b）中准确的假设了。Ray的情况相反，A点的垂直作用力向下。注意假如计算过程如上所示，那么计算的验证是有效的。
（选自Stephen P. Timoshenko and James M. Gere，材料机械，诺斯特兰德莱有限公司，1978。