过程装备与控制工程专业英语阅读材料2翻译

1、Reading Material 2 Shear 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

2、 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 str

3、esses 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

4、 the beam.The axial force ,shear force, and bending moment acting a across 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 e

5、quations of statics for the free-body diagram shown in the second part of the figure. From summations of force in the horizontal and vertical directions we find ,respectively, N=P cos V=P sin N=P cos V=P sinAnd,from a summation of moments about an axis through the centroid of cross section mn ,we ob

6、tain: M=Px sinWhere x is the distance from the free and 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

7、 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 discuss

8、ing 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 directions see Fig.1.5(a) . Therefore,we must recognize that the algebraic sign of a stress resultant does not depend

9、 upon its direction in space, such as to the left or to the right, but rather it depends 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,

10、 V and M are repeated in FIig1.6, 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 bend

11、ing 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 M,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 t

12、he middle of the beam and (b) a small distance to the right of the middle of the beam. SolutionThe first step in the analysis of this beam is to find the reactions R and R.taking moments about ends A and B gives two equations of equilibrium, from which we find R= R=+ Next ,the beam is cut at a cross

13、 section just to 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 R appear in this diagram as also do the unknown shear force V and be

14、nding moment M, both of which are shown in their positive directions ,The couple M does not appear in the figure because the beam is cut to the left of the point where Mis applied. A summation of forces in the vertical direction givesV= RP= Which shows that the shear force is negative; hence ,it act

15、s 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) givesM = Depending upon the relative magnitudes of the terms in this equation, we see that the bending moment M may be either positive or negative.To obt

16、ain 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 M now acts on the part of the beam to the left of

17、the cut section. Again summing forces in the vertical direction, and also taking moments about an axis through the cut section,we obtainV= M = +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 M,but the bending moment inc

18、rease algebraically by an amount equal to M.阅读材料2 横梁的剪切力和弯矩现在让我们细想，例如，图1.5（a）所示的悬臂梁的自由端受到一个倾斜的载荷P的作用。如果我们用一个截面mn切断横梁并隔离其左边部份作为一个自由体，如图1.5（b）所示，我们看到被隔离的部分（即右边的部分）和在其上部左边的部分保持力的平衡。在我们研究中并没有知道沿截面mn分布的应力的大小，但是我们知道这些应力的结果与载荷P达到平衡。我们很容易解决在截面上和贯穿截面质心的轴力N的作用结果，剪切力V的作用与截面平行，弯矩M作用在横梁的平面上。作用在横梁截面上的轴力，剪切力和弯矩被认为

19、是应力的作用效果。对于一个静定的横梁，应力的作用结果从平衡方程式决定。因此，对于像图1.5中的悬臂梁，我们将在本章第二部分以数字表格的形式对分离体写出三个静力学方程。从水平和垂直方向的合力中我们发现，分别是，并且，从通过截面mn质心的轴弯矩的总和，我们得到： M=Px sinX是自由端到截面mn的距离，因此，通过使用分离体的图解和静定平衡方程式，我们能够很容易地计算出应力的作用效果。横梁上的应力应归于轴力N的单独作用，这已经在第二单元被讨论过。应力N，剪切力V和弯矩M在如图1.5（b）中假设取为正向，这种习惯表示是非常有用的，当然，是在我们讨论横梁的左边分离体的力的平衡的时候。如果考虑横梁的右边部分时，在图示1.5（a）中，我们发现右边有相同大小的应力，但是方向相反。所以，我们必须认识应力的代数表示与它在空间的方向无关，正如对左边和右边部分那样，但是它依赖于材料的应力作用方向。为了阐明这个事实，惯用的标示N，V和M重复标示在了图1.6中，应力的结果展示在横梁的元素中。我们看到一个正向的轴力是从它发生作用时（拉力）的表面发射出去的，一个