外文翻譯--結(jié)構(gòu)分析的矩陣方法VIP

1 附件 1：外文資料翻譯譯文 結(jié)構(gòu)分析的矩陣方法 1. 力法和應(yīng)變方法 在前述的章節(jié)已經(jīng)介紹解決靜不定系統(tǒng)的各種各樣的方法。它們可分為兩大類(lèi)。例如，在分析拱門(mén)和框架結(jié)構(gòu)時(shí)，分析步驟如下。首先，所有的冗余的約束被對(duì)應(yīng)的冗余的力（或力矩）取代，這些力的大小可通過(guò)基于應(yīng)變能的最小勢(shì)能原理解得。類(lèi)似的過(guò)程也被用于解靜不定桁架的分析，這些方法統(tǒng)稱(chēng)為力法。 在連續(xù)梁和框架分析中，另一種不同的方法曾被使用。在這個(gè)情況下，我們首先計(jì)算了結(jié)點(diǎn)的旋轉(zhuǎn)的角度 (變形 )而冗余力是后來(lái)才求的。在連續(xù)梁的分析中使用了的 3 角度方程代表另一種方法 。這樣的方法稱(chēng)為應(yīng)變方法。 我們用一個(gè)例子來(lái)說(shuō)明這兩種方法之間的區(qū)別，如圖 10.1 的平面靜不定桁架，一力 P分解為 Px 和 PY，作用在的 5 根懸于剛性基礎(chǔ)的等截面桿交點(diǎn) A 處。因?yàn)闂U數(shù)量大于 A點(diǎn)平衡方程的數(shù)目，很明顯這是一個(gè)靜不定問(wèn)題。一般來(lái)說(shuō)，如果絞點(diǎn) A 由 n根桿鉸接而成，那么冗余的桿將是 (n-2)。因此，為了根據(jù)力法解出對(duì)應(yīng)的冗余的力 X1， X2,X3，Xn-2，我們根據(jù)這些力的作用，通過(guò)最小勢(shì)能原理獲得應(yīng)變能表達(dá)式，進(jìn)而獲得所需的方程： U/ X1=0 U/ X2=0 (a) 其中每個(gè)方程都包含所有冗余力，因此隨著桿數(shù)目的增加，方程（ a）的求解將變得越來(lái)越麻煩。 解決相同的問(wèn)題， Navier 建議使用的移置方法。在圖 10.1 的系統(tǒng)中，如果知道在力 P作用下 A 點(diǎn)的各自的水平位移 u、垂直位移 v，那么系統(tǒng)變形將完全確定下來(lái)。假設(shè) P 引 2 起的位移量很小，那么第 i桿的拉長(zhǎng)量 li=vSin ai u cosai 桿中的對(duì)應(yīng)的軸力為 Si=EAi(vSin ai u cosai)/li= EAi(vSin u cosai) in ai/h (b) 再寫(xiě)出鉸點(diǎn) A 的兩個(gè)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 從這兩個(gè) 方程中，在任一種特殊的情形下我們都很容易求出未知的 u 和 v。之后，再將 u和 v代入任何系統(tǒng)中的 (b)表達(dá)式中求出系統(tǒng)中任一根桿的 Si。對(duì)于這個(gè)問(wèn)題，可以看出，直接考慮系統(tǒng)變形使得問(wèn)題解決簡(jiǎn)單化，尤其在遇到很多根桿的時(shí)候，無(wú)需考慮桿的多少，我們只需解 2 個(gè)方程而已。 在類(lèi)似的方法下，對(duì)連續(xù)梁的直接變形分析在許多方面使問(wèn)題簡(jiǎn)單化。如果我們?nèi)コ械闹虚g支持只考慮產(chǎn)生的多余的對(duì)應(yīng)反力 X1,X2,X3，用最少勢(shì)能原理導(dǎo)出方程組 (a)，其中每個(gè)方程均包含所有的未知量。因此如果梁跨度很大，那么問(wèn)題的解決將很麻煩的。對(duì)這 個(gè)問(wèn)題的解決辦法上的重大改進(jìn)在于：將連續(xù)梁的看成兩端支撐的簡(jiǎn)單桿并計(jì)算出這根桿末端旋轉(zhuǎn)的角度。接著，根據(jù)連續(xù)梁在中間支撐處轉(zhuǎn)角一定相等的條件，已知的 3 角度方程即可獲得。這些方程比方程組 (a)簡(jiǎn)單多了，因?yàn)樗麄儧](méi)有一個(gè)包含有 3 個(gè)以上未知數(shù)。 cebdaF i g 1 0 . 2 另一個(gè)運(yùn)用應(yīng)變方法使問(wèn)題大為簡(jiǎn)單的代表例子是圖 10.2 所示系統(tǒng)。 4 個(gè)兩端固定桿剛接于 a 點(diǎn)。忽略桿中軸力影響，這個(gè)系統(tǒng)有 7 個(gè)冗余的元素，為解決這個(gè)問(wèn)題，用最少勢(shì)能原理得到 7 個(gè)方程。再用結(jié)構(gòu)應(yīng)變使問(wèn)題變得非常簡(jiǎn)單。這種變形完全是載荷作用下交點(diǎn)旋轉(zhuǎn)的角度 a 決定。解出這一角 度后，所有元素的末端可由力矩 -變形方程解出。因此，在結(jié)點(diǎn) a 的末端力矩方程的基礎(chǔ)上只需一個(gè)方程即可解出變形。 但并不能從前述討論靜不定系統(tǒng)中總結(jié)出應(yīng)變方法總比力法要優(yōu)異。例如，在一個(gè)含 3 有 1 個(gè)冗余度和 10 個(gè)結(jié)點(diǎn)的簡(jiǎn)單桁架中，用上面的應(yīng)變的方法將變得很麻煩，而使用的力法是極其簡(jiǎn)單的。 在處理高次靜不定系統(tǒng)時(shí)，我們通常發(fā)現(xiàn)那不管我們用的力法還是應(yīng)變方法，都要解帶有許多未知量的線性代數(shù)方程組。拋開(kāi)結(jié)構(gòu)分析的其他任何特別的問(wèn)題，讓我們考慮如下系統(tǒng)的方程： 1121211 cxaxaa nn 2222221 cxaxaa nn . （ 10.1） mnmnmm cxaxaa 221 理論上講，這種線性代數(shù)方程總是可解的，但是隨著方程數(shù)目的增加，解方程的過(guò)程將變得十分麻煩，為了簡(jiǎn)化解題技巧，介紹一種矩陣代數(shù)的記法。因此，在矩陣記法中，方程 (10.1) 可精簡(jiǎn)為： aijxj=ci （ 10.1a） 或簡(jiǎn)記 Ax=c （ 10.1b） 方括號(hào)表達(dá)式中的每個(gè)數(shù)組 (或記法 )被稱(chēng)為一個(gè)矩陣。數(shù)（或記法 )本身被稱(chēng)為元素，當(dāng)矩陣有 m 行和 n 列時(shí)，矩陣被稱(chēng)為 m*n 型。當(dāng)僅僅在矩陣有一列或一行元素時(shí)，它被稱(chēng)為列向量或行向量。認(rèn)為（ 10.1a）矩陣 aij以這種方式作用于列向量 xj組成了上面方程組的左邊。因此有必要去學(xué)習(xí)一些矩陣代數(shù)的規(guī)則。 但在這之前，讀者應(yīng)認(rèn)清結(jié)構(gòu)分析的矩陣方法并沒(méi)有什么特別的或不可思議的，也并不代表它比前述章節(jié)討論的手算 方法更為優(yōu)越。它真正的優(yōu)勢(shì)在于它引導(dǎo)去更好的利用了電子計(jì)算機(jī)。因此，避免了棘手的手算麻煩而另辟了一條結(jié)構(gòu)分析的道路。在可得到的有限的空間里，我們將不可能揭露矩陣方法的全部作用，但通過(guò)簡(jiǎn)單的例子幫助讀者熟悉方法并領(lǐng)會(huì)他的優(yōu)點(diǎn)。 2 連續(xù)結(jié)構(gòu)的矩陣分析方法 諸如建筑結(jié)構(gòu)的連續(xù)結(jié)構(gòu)很可能是高次靜不定的，以致于在分析時(shí)要處理分析許多未知數(shù)。解決這類(lèi)問(wèn)題的唯一的可行方法是求助于電子數(shù)字計(jì)算機(jī)。并且為實(shí)現(xiàn)這個(gè)目的，矩陣陳述是最有利的。為闡述這類(lèi)問(wèn)題的矩陣方法，我們以圖（ 10.13）的二層結(jié)構(gòu)框架來(lái) 4 舉例說(shuō)明，盡管這個(gè) 框架并沒(méi)有使問(wèn)題復(fù)雜的眾多未知數(shù)，但在另一方面，它足以闡述清涵蓋分析更大結(jié)構(gòu)時(shí)所有的步驟、過(guò)程。 為簡(jiǎn)潔起見(jiàn)，我們假設(shè)每段梁的長(zhǎng)為 l，一樣的彎曲剛度 EI，因此硬度條件都是相等的，即 k=EI/l是一樣的。作為一個(gè)一般練習(xí)，忽略軸應(yīng)力和剪應(yīng)力引起的變形，而僅僅考慮彎曲變形。在這些假設(shè)前提下，在負(fù)載作用下的結(jié)構(gòu)的變形完全由 6 個(gè)位移量決定。即，兩個(gè)水平位移 a， b 四個(gè)交點(diǎn)處的旋轉(zhuǎn)角度 1, 2， 3, 4。 6 個(gè)位移量求出來(lái)以后，所有末端力矩可通過(guò)力位移方程計(jì)算出，這個(gè)問(wèn)題就解決了。因此，我們介紹列向量 j= a， b， 1, 2, 3, 4 (a) 并將這一系列位移量作為問(wèn)題未知量。 圖 10.13 作為計(jì)算位移量的第一步，我們首先考慮圖 10.14 舉例說(shuō)明了的 2 個(gè)簡(jiǎn)單的問(wèn)題。在圖 10.14a 中，在兩端固定的等截面梁 AB 的端點(diǎn) A 作用一個(gè)位移， A 沒(méi)有任何旋轉(zhuǎn)運(yùn)動(dòng)，B 沒(méi)有任何移動(dòng)。那么， A、 B 兩點(diǎn)的反力根據(jù)方程（ 9.6）很容易就計(jì)算出了。并且我們發(fā)現(xiàn) Rab=12k/l2 Mab=6k/l Rab=12k/l2 Mab=6k/l (b) 在圖 10.14b 中，相同梁的端點(diǎn) A 只有一個(gè)旋轉(zhuǎn)角度，不允許 A 有任何側(cè)面移動(dòng)，端點(diǎn) B 也沒(méi)有任何移動(dòng)。接著，再使用應(yīng)力 -變形方程 9.6，我們發(fā)現(xiàn) Rab=6k/l Mab=4k Rab=6k/l Mab=2k (b) 5 圖 10.14 在方程 (b)和 (b)中，出現(xiàn)在和前面的系數(shù)代表梁端部的反力、力或約束，而此時(shí)位移和都是單位位移。對(duì)應(yīng)于梁中每一種類(lèi)型的位移的量被稱(chēng)為剛度影響系數(shù)。為了參考便利，這些剛度影響系數(shù)以矩陣形式標(biāo)注圖 10.14 的每根橫梁下面。 現(xiàn)在，讓我們回到圖 10.13 的結(jié)構(gòu)中，移去所有的已加負(fù)載，并且并交點(diǎn)處無(wú)傳遞和旋轉(zhuǎn)。完了后，我們移開(kāi)與系統(tǒng) 6 個(gè)自由度之一相對(duì)應(yīng)的任一約束，叫約束 j，并給予單位位移 j=1。這將導(dǎo)致與這個(gè)人為約束相一致的結(jié)構(gòu)變形，接著我們計(jì)算出 6 個(gè)自由度對(duì)應(yīng)的其余結(jié)果。那就是說(shuō)，在假定 j=1 的情況下，計(jì)算出了支持結(jié)構(gòu)系統(tǒng)所需要的外力和外力偶。總的來(lái)說(shuō)，在 i處的反力不管是外力還是 外力偶，我們都標(biāo)記為外反應(yīng) Sij，因此，剛度影響系數(shù) Sij 定義為在在 j 處作用單位位移，其他位移均為 0 的情況下所需施加的外力。在這個(gè)例子中將有 36 個(gè)這些剛度影響系數(shù)，我們現(xiàn)在利用圖 10.14 所示的單根桿的剛度影響系數(shù)完成整個(gè)系統(tǒng)（剛度影響系數(shù)）的計(jì)算。 在圖 10.15a 中，在單位位移 a=1 時(shí)，即最高的地板的側(cè)面的位移為一個(gè)單位移，所有的另外的位移均相等為零。那么，支撐結(jié)構(gòu)所要求的外部力標(biāo)注在圖中，并且其大小也列在結(jié)構(gòu)旁邊。在這些計(jì)算中，我們規(guī)定線形位移和力向右為正，向左為負(fù)，角位移順時(shí)針?lè)较驗(yàn)檎磿r(shí)針?lè)较?為負(fù)。例如， Sba 的計(jì)算，見(jiàn)圖 10.14a，我們看到每個(gè)頂層列的底部的反力。圖 10.15a 左部有 2 個(gè)如此的列并且（結(jié)果）是 12k/l2；因此，圖上標(biāo)注Sba=-24k/l2。再考慮 S4a 的計(jì)算 .由圖 10.14a 的結(jié)果，圖 10.15a，列的 4a的反作用力矩是反時(shí)針?lè)较虻?，其大小?6k/l并且僅僅有一列；因此， S4a = -6k/l。讀者應(yīng)該自己檢查其他的 Sij 的值。 6 下一步，在圖 10. 15b 中，設(shè)單位位移 b= 1，即中間層的單元的一個(gè)單位水平位移，其他位移均相等為零。那么，同上方法，使用圖 10.14a 的剛度影響系數(shù)。求出外反力 Sij見(jiàn)圖所示。與應(yīng)變模式 1=1, 2=1, 3=1, 4=1 對(duì)應(yīng)的誘導(dǎo)外力被標(biāo)注在圖 10.15c， d， e，f，這就完成了整個(gè)結(jié)構(gòu)的影響系數(shù)的計(jì)算。 現(xiàn)在就將這些剛度系數(shù)集中成方陣格式，叫做結(jié)構(gòu)剛度矩陣。行和列都按 a， b， 1， 2，3， 4 的順序?qū)懗?。那就成?可以觀察到這是一個(gè)對(duì)稱(chēng)矩陣，并且這種對(duì)稱(chēng)來(lái)自于協(xié)調(diào)理論的 有了上面矩陣 (c)所示剛度影響系數(shù)以后，我們可以利用重疊原則計(jì)算出任何數(shù)據(jù)組合位移 j 的條件下支持框架結(jié)構(gòu)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，我們正在尋找那些在圖 10.13 系統(tǒng)所示的外力作用下的位移的一系列值，那些力是是實(shí)實(shí)在在的結(jié)構(gòu)負(fù)載。真實(shí)的位移集合已在系統(tǒng)的代數(shù)方程中定義了 其中符號(hào)相反的 ql2 /12 和 -ql2 /12 表示梁 34 的端部力矩，即結(jié)點(diǎn) 3 和 4 各自的不平衡力矩。介紹矩陣記法 Fi=Pa Pb 0 Qa ql2/12 -ql2/12 (d) 它被稱(chēng)為負(fù)載矩陣， 例 (I0.17)的矩陣記為 Sij j = Fi (10.17a) 在這個(gè)方程出現(xiàn)的 3 個(gè)矩陣各自表達(dá)為 (c), (a)和 (d)。 例 (I0.17a)，方程位移的解為 i = Sij- 1 Fi 我們注意到求解需要?jiǎng)偠染仃?Sij的逆矩陣， 這時(shí)我們就需要計(jì)算機(jī)的幫助了。 8 附件 2：外文資料翻譯原文 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