外文翻译--结构分析的矩阵方法

1 附件 1：外文资料翻译译文 结构分析的矩阵方法 1. 力法和应变方法 在前述的章节已经介绍解决静不定系统的各种各样的方法。它们可分为两大类。例如，在分析拱门和框架结构时，分析步骤如下。首先，所有的冗余的约束被对应的冗余的力（或力矩）取代，这些力的大小可通过基于应变能的最小势能原理解得。类似的过程也被用于解静不定桁架的分析，这些方法统称为力法。 在连续梁和框架分析中，另一种不同的方法曾被使用。在这个情况下，我们首先计算了结点的旋转的角度 (变形 )而冗余力是后来才求的。在连续梁的分析中使用了的 3 角度方程代表另一种方法 。这样的方法称为应变方法。 我们用一个例子来说明这两种方法之间的区别，如图 10.1 的平面静不定桁架，一力 P分解为 Px 和 PY，作用在的 5 根悬于刚性基础的等截面杆交点 A 处。因为杆数量大于 A点平衡方程的数目，很明显这是一个静不定问题。一般来说，如果绞点 A 由 n根杆铰接而成，那么冗余的杆将是 (n-2)。因此，为了根据力法解出对应的冗余的力 X1， X2,X3，Xn-2，我们根据这些力的作用，通过最小势能原理获得应变能表达式，进而获得所需的方程： U/ X1=0 U/ X2=0 (a) 其中每个方程都包含所有冗余力，因此随着杆数目的增加，方程（ a）的求解将变得越来越麻烦。 解决相同的问题， Navier 建议使用的移置方法。在图 10.1 的系统中，如果知道在力 P作用下 A 点的各自的水平位移 u、垂直位移 v，那么系统变形将完全确定下来。假设 P 引 2 起的位移量很小，那么第 i杆的拉长量 li=vSin ai u cosai 杆中的对应的轴力为 Si=EAi(vSin ai u cosai)/li= EAi(vSin u cosai) in ai/h (b) 再写出铰点 A 的两个平衡方程， 得 v Ai Sin2 ai Cos ai-u Ai Cos2aiSin ai =Pxh/E (c) v Ai Sin2 ai-u Ai Sin2 ai Cos ai=Pyh/E 从这两个 方程中，在任一种特殊的情形下我们都很容易求出未知的 u 和 v。之后，再将 u和 v代入任何系统中的 (b)表达式中求出系统中任一根杆的 Si。对于这个问题，可以看出，直接考虑系统变形使得问题解决简单化，尤其在遇到很多根杆的时候，无需考虑杆的多少，我们只需解 2 个方程而已。 在类似的方法下，对连续梁的直接变形分析在许多方面使问题简单化。如果我们去除所有的中间支持只考虑产生的多余的对应反力 X1,X2,X3，用最少势能原理导出方程组 (a)，其中每个方程均包含所有的未知量。因此如果梁跨度很大，那么问题的解决将很麻烦的。对这 个问题的解决办法上的重大改进在于：将连续梁的看成两端支撑的简单杆并计算出这根杆末端旋转的角度。接着，根据连续梁在中间支撑处转角一定相等的条件，已知的 3 角度方程即可获得。这些方程比方程组 (a)简单多了，因为他们没有一个包含有 3 个以上未知数。 cebdaF i g 1 0 . 2 另一个运用应变方法使问题大为简单的代表例子是图 10.2 所示系统。 4 个两端固定杆刚接于 a 点。忽略杆中轴力影响，这个系统有 7 个冗余的元素，为解决这个问题，用最少势能原理得到 7 个方程。再用结构应变使问题变得非常简单。这种变形完全是载荷作用下交点旋转的角度 a 决定。解出这一角 度后，所有元素的末端可由力矩 -变形方程解出。因此，在结点 a 的末端力矩方程的基础上只需一个方程即可解出变形。 但并不能从前述讨论静不定系统中总结出应变方法总比力法要优异。例如，在一个含 3 有 1 个冗余度和 10 个结点的简单桁架中，用上面的应变的方法将变得很麻烦，而使用的力法是极其简单的。 在处理高次静不定系统时，我们通常发现那不管我们用的力法还是应变方法，都要解带有许多未知量的线性代数方程组。抛开结构分析的其他任何特别的问题，让我们考虑如下系统的方程： 1121211 cxaxaa nn 2222221 cxaxaa nn . （ 10.1） mnmnmm cxaxaa 221 理论上讲，这种线性代数方程总是可解的，但是随着方程数目的增加，解方程的过程将变得十分麻烦，为了简化解题技巧，介绍一种矩阵代数的记法。因此，在矩阵记法中，方程 (10.1) 可精简为： aijxj=ci （ 10.1a） 或简记 Ax=c （ 10.1b） 方括号表达式中的每个数组 (或记法 )被称为一个矩阵。数（或记法 )本身被称为元素，当矩阵有 m 行和 n 列时，矩阵被称为 m*n 型。当仅仅在矩阵有一列或一行元素时，它被称为列向量或行向量。认为（ 10.1a）矩阵 aij以这种方式作用于列向量 xj组成了上面方程组的左边。因此有必要去学习一些矩阵代数的规则。 但在这之前，读者应认清结构分析的矩阵方法并没有什么特别的或不可思议的，也并不代表它比前述章节讨论的手算 方法更为优越。它真正的优势在于它引导去更好的利用了电子计算机。因此，避免了棘手的手算麻烦而另辟了一条结构分析的道路。在可得到的有限的空间里，我们将不可能揭露矩阵方法的全部作用，但通过简单的例子帮助读者熟悉方法并领会他的优点。 2 连续结构的矩阵分析方法 诸如建筑结构的连续结构很可能是高次静不定的，以致于在分析时要处理分析许多未知数。解决这类问题的唯一的可行方法是求助于电子数字计算机。并且为实现这个目的，矩阵陈述是最有利的。为阐述这类问题的矩阵方法，我们以图（ 10.13）的二层结构框架来 4 举例说明，尽管这个 框架并没有使问题复杂的众多未知数，但在另一方面，它足以阐述清涵盖分析更大结构时所有的步骤、过程。 为简洁起见，我们假设每段梁的长为 l，一样的弯曲刚度 EI，因此硬度条件都是相等的，即 k=EI/l是一样的。作为一个一般练习，忽略轴应力和剪应力引起的变形，而仅仅考虑弯曲变形。在这些假设前提下，在负载作用下的结构的变形完全由 6 个位移量决定。即，两个水平位移 a， b 四个交点处的旋转角度 1, 2， 3, 4。 6 个位移量求出来以后，所有末端力矩可通过力位移方程计算出，这个问题就解决了。因此，我们介绍列向量 j= a， b， 1, 2, 3, 4 (a) 并将这一系列位移量作为问题未知量。 图 10.13 作为计算位移量的第一步，我们首先考虑图 10.14 举例说明了的 2 个简单的问题。在图 10.14a 中，在两端固定的等截面梁 AB 的端点 A 作用一个位移， A 没有任何旋转运动，B 没有任何移动。那么， A、 B 两点的反力根据方程（ 9.6）很容易就计算出了。并且我们发现 Rab=12k/l2 Mab=6k/l Rab=12k/l2 Mab=6k/l (b) 在图 10.14b 中，相同梁的端点 A 只有一个旋转角度，不允许 A 有任何侧面移动，端点 B 也没有任何移动。接着，再使用应力 -变形方程 9.6，我们发现 Rab=6k/l Mab=4k Rab=6k/l Mab=2k (b) 5 图 10.14 在方程 (b)和 (b)中，出现在和前面的系数代表梁端部的反力、力或约束，而此时位移和都是单位位移。对应于梁中每一种类型的位移的量被称为刚度影响系数。为了参考便利，这些刚度影响系数以矩阵形式标注图 10.14 的每根横梁下面。 现在，让我们回到图 10.13 的结构中，移去所有的已加负载，并且并交点处无传递和旋转。完了后，我们移开与系统 6 个自由度之一相对应的任一约束，叫约束 j，并给予单位位移 j=1。这将导致与这个人为约束相一致的结构变形，接着我们计算出 6 个自由度对应的其余结果。那就是说，在假定 j=1 的情况下，计算出了支持结构系统所需要的外力和外力偶。总的来说，在 i处的反力不管是外力还是 外力偶，我们都标记为外反应 Sij，因此，刚度影响系数 Sij 定义为在在 j 处作用单位位移，其他位移均为 0 的情况下所需施加的外力。在这个例子中将有 36 个这些刚度影响系数，我们现在利用图 10.14 所示的单根杆的刚度影响系数完成整个系统（刚度影响系数）的计算。 在图 10.15a 中，在单位位移 a=1 时，即最高的地板的侧面的位移为一个单位移，所有的另外的位移均相等为零。那么，支撑结构所要求的外部力标注在图中，并且其大小也列在结构旁边。在这些计算中，我们规定线形位移和力向右为正，向左为负，角位移顺时针方向为正，反时针方向 为负。例如， Sba 的计算，见图 10.14a，我们看到每个顶层列的底部的反力。图 10.15a 左部有 2 个如此的列并且（结果）是 12k/l2；因此，图上标注Sba=-24k/l2。再考虑 S4a 的计算 .由图 10.14a 的结果，图 10.15a，列的 4a的反作用力矩是反时针方向的，其大小为 6k/l并且仅仅有一列；因此， S4a = -6k/l。读者应该自己检查其他的 Sij 的值。 6 下一步，在图 10. 15b 中，设单位位移 b= 1，即中间层的单元的一个单位水平位移，其他位移均相等为零。那么，同上方法，使用图 10.14a 的刚度影响系数。求出外反力 Sij见图所示。与应变模式 1=1, 2=1, 3=1, 4=1 对应的诱导外力被标注在图 10.15c， d， e，f，这就完成了整个结构的影响系数的计算。 现在就将这些刚度系数集中成方阵格式，叫做结构刚度矩阵。行和列都按 a， b， 1， 2，3， 4 的顺序写出。那就成为 可以观察到这是一个对称矩阵，并且这种对称来自于协调理论的 有了上面矩阵 (c)所示刚度影响系数以后，我们可以利用重叠原则计算出任何数据组合位移 j 的条件下支持框架结构所需的外力。例如，要求外力是 Fa= Saaa+ Sabb+ Sa11+ Sa22+ Sa33+ Sa44 要求外力偶是 7 M1 = S1a a+ S1b b+ S11 1+ S12 2+ S13 3+ S14 4 等等。然而，我们正在寻找那些在图 10.13 系统所示的外力作用下的位移的一系列值，那些力是是实实在在的结构负载。真实的位移集合已在系统的代数方程中定义了 其中符号相反的 ql2 /12 和 -ql2 /12 表示梁 34 的端部力矩，即结点 3 和 4 各自的不平衡力矩。介绍矩阵记法 Fi=Pa Pb 0 Qa ql2/12 -ql2/12 (d) 它被称为负载矩阵， 例 (I0.17)的矩阵记为 Sij j = Fi (10.17a) 在这个方程出现的 3 个矩阵各自表达为 (c), (a)和 (d)。 例 (I0.17a)，方程位移的解为 i = Sij- 1 Fi 我们注意到求解需要刚度矩阵 Sij的逆矩阵， 这时我们就需要计算机的帮助了。 8 附件 2：外文资料翻译原文 Matrix methods in structural analysis 1 FORCE AND DEFORMATION METHODS The various methods of analysis of statically indeterminate systems that have been used in preceding chapters fall into two distinct classifications. In the analysis of arches and frames, for example ,the procedure was as follows: First, all redundant constraints were removed and replaced by the corresponding redundant forces(or moments).The magnitudes of these forces were then found by using the theorem of least work based on a consideration of the strain energy in the structure. A similar procedure was used in the analysis of statically indeterminate trusses. This general approach is called the method of forces. In the analysis of continuous beams and frames, a somewhat different procedure was used. In this case, we calculated first the angles of rotation of the joints (deformations) and considered the redundant forces only later. The three-angle equation used in the analysis of continuous beams represents again the kind of approach. Such procedure is called the method of deformation. To illustrate, on the same example, the distinction between the two methods, let us consider the statically indeterminate plane truss shown in Fig 10.1. Here, a load P, defined by its components Px and Py, is supported by five prismatic members hinged together at A and to a rigid foundation at their upper ends, Since the number of bars is greater than the number of equations of equilibrium for the joint A, the problem is evidently statically indeterminate . In general, if the hinge A is attached to the foundation by n bars, all in one plane, the number of redundant bars will be (n-2). Then, to determine the corresponding redundant forces X1,X2,X3, ,Xn-2 by a method of forces, we write the expression for the strain energy of the system as a function of these forces and, by using the theorem of least work, obtain the necessary equations: U/ X1 U/ X2 (a) Each of these equations will contain all of the redundant forces, so with the increase in the number of bars, the solution of Esq.(a) becomes more and more cumbersome. To solve the same problem, Nervier suggested the use of a method of displacements. The deformation of the system in Fig 10.1 is completely determined if we know the horizontal and vertical components u and v, respectively, of the displacement of the hinge A produced by the load P. Assuming that these displacements are small, the elongation of any bar i will then be li=v Sinai u cosai And the corresponding axial force in the bar becomes Si=EAi(v Sinai u cosai)/li= EAi(v Sinai u cosai) Sinai/h (b) Writing now the two equations of equilibrium for the hinge A, we obtain v Ai Sin2 ai Cos aiu Ai Cos 2 ai Sin ai =Pxh/E (c) v Ai Sin2 aiu Ai Sin2 ai Cos ai=Pyh/E 9 From these two equations, the unknowns u and v can be readily calculated in each particular case. After this, substitution of u and v into expression (b) gives us the force Si in any bar of the system. It is seen that for this problem, direct consideration of the deformation of the system results in a substantial simplification of the solution, especial if there are a large number of bars since, independently of that number, we have to solve only two equations. In a similar way, direct consideration of the deformations simplifies the analysis of a continuous beam on many supports. If we remove all intermediate supports and consider the corresponding reactions X1,X2,X3, as the redundant quantities, the theorem of least work yields a system of equations(a),each of which contains all of the unknowns. Thus, the solution of the problem becomes very cumbersome if the number of spans is large. A great improvement in the solution of this problem is attained by considering each span of the continuous beam as a simple beam on two supports and calculating the angles of rotation of the ends of such beams. Then, from the condition that at each intermediate support these angles for two adjacent spans must be equal, the known three-angle equations are obtained. Such equations are much simpler than Esq.(a) because no one of them contains more than three unknowns. cebdaF i g 1 0 . 2 Another example in which the method of deformations resulted in a great simplification is represented by the system shown in Fig10. 2, where four members are rigidly joined together at a and built in at their far ends. Neglecting the effect of axial forces in the bars, this system has seven redundant reactive elements, and for their determination, the theorem of least work would give seven equations. Again, the problem was greatly simplified by considering the deformation of the structure. This deformation is completely def ined by the angle of rotation a of the joint a produced by the applied loads. When the magnitude of this angle is found。The end moments for all the members can be readily calculated from the slope-deflection equations. Thus, by considering deformations first, we need only one equation which was written on the basis of the equilibrium of the end moments at joint a. It is not to be concluded from the foregoing discussion that, in the analysis of a statically indeterminate system, a method of deformations is always superior to a method of forces. For example, in the case of a simple truss having one redundant reaction and ten joints The method of deformations described above would become very cumbersome, whereas the method of forces used is extremely simple. In dealing with highly statically indeterminate systems, we usually find that regardless of whether we use a method of forces or a method of deformations, it becomes necessary to solve a large number of simultaneous linear algebraic equations with as many unknowns Without regard to any particular problem of structural analysis, let us now consider a system of such equations: 1121211 cxaxaa nn 10 2222221 cxaxaa nn . mnmnmm cxaxaa 221 Theoretically, such a system of linear algebraic equations can always be solved, but the progress of solution becomes cumbersome as the number of equations increases, and to simplify the technique of this solution, the notation of matrix algebra will now be introduced. Thus, in matrix notations, Eqs.(10.1) may be written in the condensed form aijxj=ci （ 10.1a） Or simply Ax=c （ 10.1b） Each array of numbers(or symbols) in the brackets of expression is called a matrix. The numbers (or symbols) themselves are called elements, and when there are m rows and n columns, the matrix is said to be of order m*n. When there is only one column or one row of elements in the matrix, it is called a column vector or a row vector. It is understood that the matrix aij in (10.1a) operates or the column vector xj in such a way as to produce the left-hand side of the system of equations above. This brings us to the necessity to learn some of the rules of matrix algebrathods. Before proceeding with this, however, the reader should understand that the use of matrix methods in structural analysis holds no particular magic, nor does it represent any great advantage over the methods discussed in preceding chapters so long as numerical calculations are to be made by hand. Its real advantage lies in the fact that it lends itself particularly well to the use of the electronic digital computer and thereby opens the door to the analysis of structural problems that would otherwise be too involved and complex to cope with by desk-calculator techniques. In the limited space available here, we shall be unable to disclose the full power of the matrix approach, but it is hoped that the simple examples to be discussed will give the reader enough familiarity with the method to. enable him to study the literature on tile subject to better advantage. 2.MATRIX ANALYSIS OF CONTINUOUS FRAMES Continuous frame structures such as building frames are likely to be highly statically indeterminate so that in their analysis we have to deal with a large number of unknowns. The only practicable way of solving such problems is to have recourse to the electronic digital computer, and for this purpose a matrix formulation of the problem is the most advantageous. To illustrate a matrix method for such problems, we shall consider here a two-story building frame as shown in Fig.（ 10.l 3） On the one hand, this frame will not involve so many unknowns as to make the discussion unwieldy, yet, on the other hand, it will be extensive enough to permit us to illustrate all the steps that would be required in the analysis of a much larger structure. For simplicity, we assume that each member has the same length l and the same flexural rigidity EI so that the stiffness factors are all equal, that is, k = EI/l is the same for all members. As is usual practice, we also neglect the deformations caused by axial forces and by shearing forces in the members and consider only bending deformation. Under these assumptions, the deformation of the frame under load will be completely defined by a set of six displacements: namely, the horizontal displacements a , and b of the two floors and the angles of rotation 1, 2， 3, 4 of the four rigid joints. When these six displacements have been found, all end moments can be calculated from the slope-deflection equations , and the problem is solved. We therefore introduce the column vector 11 j = a, b, 1, 2, 3, 4 (a) and select this set of displacements as the unknowns of the problem. Fig10.13 As a preliminary step to the calculation of the displacements , we first consider the two simple problems illustrated in Fig. 10.14. In Fig. 10.14a, we give to the end A of a prismatic beam AB with built-in ends a displacement , without allowing any rotation of the tangent at A or any movement at all of the end B. Then, the reactions at A and B can easily be calculated by using the slope-deflection equations (9.6)and we find Rab=12k /l2 Mab=6k /l Rab=12k /l2 Mab=6k /l (b) In Fig. 10.14b, the end A of the same beam is given an angle of rotation without allowing any lateral deflection of A or any movement at all of the B. Then, again using the slope-deflection equations 9.6 we find Rab=6k /l Mab=4k Rab=6k /l Mab=2k (b) The coefficients appearing in front of and in Fqs. (b) and (b) are seen to represent the reactions, or forces of constraint, at the ends of the beam when the displacements and are each equal to unity. These quantities are called the stiffness influence coefficients for the beam corresponding to each type of displacement. For convenience of easy reference, these stiffness coefficients are recorded in matrix form under each beam in Fig. 10.14. , Now, let us return to the frame in Fig. 10.13, remove all applied loads, and lock all joints against both translation and rotation.This done, we remove just one constraint corresponding to any one of the six degrees of freedom of the system, say constraint j, and make there a unit displacement j=1. This will result in some deformation of the structure consistent with the remaining artific ial constraints, and we proceed to calculate the reaction corresponding to each of the six degrees of freedom. That is, we calculate the system of external forces and couples necessary to hold the structure in the assumed configuration defined by j=1. In the generalized sense, we denote such an external reaction at i by Sij regardless of whether it is a force or a couple. 12 Thus, we define the stiffness influence coefficient Sij as the external reaction at i due to an imposed unit displacement at j when all other displacements are held equal to zero. In our example there will be 36 of these stiffness influence coefficients, and we now set about their calculation, making use of the single-member stiffness coefficients shown in Fig. 10.14. Let us begin in Fig. 10.15a with a unit displacement a=1, that is, a unit lateral displacement of the top floor, all other displacements being held equal to zero. Then, the external forces required to hold the structure in this configuration will act as shown in the figure, and their magnitudes will be as listed beside the structure. In these calculations,we consider linear displacements and forces positive to the right, negative to the left, and angular displacements and couples positive when clockwise, negative when counterclockwise. Consider, for example, the calculation of Sba. From Fig. 10.14a, we see that the reaction at the- bottom of each upper-story column in Fig. 10.15a is 12k/l2 acting to the left and that there are two such columns； hence, Sba=-24k/l2 as shown. Consider, again, the calculation of S4a. From Fig. 10.14a, we see that the reactive moment at the bottom of the column 42 in Fig. 10.15a is counterclockwise and of magnitude 6k/l and that there is only one column； hence, S4a = -6k/l. The reader should check tile other values of Sij for himself. Next, in Fig. 10. 15b, we make a unit displacement b = 1, that is, a unit horizontal displacement of the middle floor, holding all other displacements equal to zero. Then, as before, using the stiffness coefficients from Fig. 10. 14a, we find the external reactions Sij as s