外文翻譯--結(jié)構(gòu)分析的矩陣方法

1、南 京 理 工 大 學(xué)畢業(yè)設(shè)計(jì)(論文)外文資料翻譯學(xué)院（系）：機(jī)械工程學(xué)院專姓學(xué)業(yè)：名：號(hào)：機(jī)械工程及 自動(dòng)化徐峰0101500131外文出處：( 用外文寫 )theory of structurespublisher:mcgraw hill附件： 1.外文 資料翻 譯譯 文； 2.外文原文。指導(dǎo)教師評(píng)語 ：翻譯內(nèi)容符合 畢業(yè)設(shè)計(jì)內(nèi)容的要求，翻譯工作量較大，翻譯基本正確、 符合科技外語的翻譯習(xí)慣和用法，較好的完成了翻譯工作。簽名：年月日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。例如，在分析拱門

2、 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基于應(yīng)變能的最小勢(shì)能原理解得。類似的過程也被用于解靜 不定桁架的分析，這些方法統(tǒng)稱為力法。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。這樣的方法稱為應(yīng)變方法。我們用一個(gè)例子 來說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

3、是一個(gè)靜不定問題。一般來說，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=0u/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）的求解將變得越來越麻煩。解決相同的問題 ，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。假設(shè)p2引起的位移量很 小，那么第 i 桿的拉長量li=vsin aiu

4、cosai桿中的對(duì)應(yīng)的軸 力為si=eai(vsinaiu cosai)/li= eai(vsin u cosai) in ai/h(b)再寫出鉸點(diǎn) a 的兩個(gè)平衡方程，得2v ai sin ai cosai-u ai cos2aisin ai =pxh/e (c)2 2v ai sin ai-u ai sin 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)問題，可以看出，直接考慮系統(tǒng)變 形使得問題解決簡單化，尤其在遇到很多根桿的時(shí)候，無需考慮桿的多

5、少，我們只需解 2 個(gè)方程而已。在類似的方法下 ，對(duì)連續(xù)梁的直接變形分析在許多方面使問題簡單化。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，那么問題的解決將很麻煩的。對(duì) 這個(gè)問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)簡單多了，因?yàn)樗麄儧]有一個(gè)包含有 3 個(gè)以上 未知數(shù)。eadbcfig 10.2另一個(gè)運(yùn)用應(yīng)變 方法使問題大為簡單的代

6、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)變使問題變得非常簡單。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)3含有 1 個(gè)冗余度 和 10 個(gè)結(jié)點(diǎn)的簡單 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 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

7、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。拋開結(jié)構(gòu)分析的其他任何特別的問題，讓我們考慮如下系統(tǒng)的方 程：a11a12x2a1nxnc1a21 a22x2a2nxnc2 .（10.1）am1am2x2amnxncm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，解方程的過程將變得十分麻 煩，為了簡化解題技巧，介紹一種矩陣代數(shù)的記法。因此，在矩陣記法中，方程 (10.1) 可精簡為：aijxj=ci（10.1a）或簡記（10.1b）ax=c方括號(hào)表達(dá)式中 的每個(gè)數(shù)組 (或記法 )被稱為一個(gè)矩陣 。數(shù)（或記法 )本身被稱為元素，當(dāng)矩陣有 m 行和 n 列時(shí)，矩

8、陣被稱為m*n 型。 當(dāng)僅僅在矩陣有一列或一行元素時(shí)，它被稱為列向量或行向量。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，但通過簡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

9、不定的，以致于在分析時(shí)要處理分析許多4未知數(shù)。解決這 類問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。為闡述這類問題的矩陣方法，我們以圖（10.13）的二層結(jié)構(gòu)框架來舉例說明 ，盡管這個(gè)框架并沒有使問題復(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、過程。為簡潔起見，我 們假設(shè)每段梁的長為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

10、個(gè)交點(diǎn)處的旋轉(zhuǎn)角度1,2，3,4。6 個(gè)位移量求出來以后，所有末 端力矩可通過力位移方程計(jì)算出，這個(gè)問題就解決了。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 舉例說明了的 2 個(gè)簡單的問題。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 沒有任何旋轉(zhuǎn)運(yùn)動(dòng)， b 沒有任何 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/l2mab=6k/l(b) 在圖 10.1

11、4b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)于梁中每一種類型的位移的量被稱為剛度影響系數(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?，F(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)處無傳遞和旋轉(zhuǎn)。完了后， 我們移開與系統(tǒng) 6 個(gè)自由度之一相

12、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，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?？?的來說，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

13、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)針方向?yàn)?正，反時(shí)針方向?yàn)樨?fù)。例如，sba 的計(jì)算，見圖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)針方 向的，其大小為 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

14、。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?，F(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?。那就成為可以觀察到這是 一個(gè)對(duì)稱矩陣，并且這種對(duì)稱來自于協(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

15、框架結(jié)構(gòu)所需的外力。例如，要求外力是fa=saa a+ sabb+ sa1 1+ sa22+ sa33+ sa4 4要求外力偶是7m1 = s1aa+ s1bb+ s111+ s122+ s133+ s144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2 2其中符號(hào)相反的 ql /12 和-ql /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=papb 0 qa ql2/12-ql2/12(d)它被稱為負(fù)載矩 陣， 例 (i0.17)

16、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 = sijfi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- 1附件 2：外 文資料翻譯原文matrix methods in structural analysis1 force and deformationmethodsthe various methods of analysis of statically indeterminate systems that have been used in

17、precedingchapters fall into two distinct classifications. in the analysis of arches and frames, for example ,the procedurewas as follows: first, all redundant constraints were removed and replaced by the corresponding redundantforces(or moments).the magnitudes of these forces were then found by usin

18、g the theorem of least work basedon a consideration of the strain energy in the structure. a similar procedure was used in the analysis ofstatically indeterminate trusses. this general approach is called the method of forces.in the analysis of continuous beams and frames, a somewhat different proced

19、ure was used. in this case,we calculated first the angles of rotation of the joints (deformations) and considered the redundant forces onlylater. 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

20、.to illustrate, on the same example, the distinction between the two methods, let us consider the staticallyindeterminate plane truss shown in fig 10.1. here, a load p, defined by its components px and py, is supportedby five prismatic members hinged together at a and to a rigid foundation at their

21、upper ends, since the numberof bars is greater than the number of equations of equilibrium for the joint a, the problem is evidentlystatically 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 de

22、termine the corresponding redundant forcesx1,x2,x3, ,xn-2 by a method of forces, we write the expression for the strain energy of the system as afunction 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

23、 the redundant forces, so with the increase in the number of bars, the solution of esq.(a) becomesmore 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 a

24、nd vertical components u and v, respectively, of the displacement of the hinge a produced by the load p. assuming that these displacementsare small, the elongation of any bar i will then beli=v sinai u cosaiand the corresponding axial force in the bar becomes si=eai(vsinai u cosai)/li= eai(v sinai u

25、 cosai)sinai/h (b)writing now the two equations of equilibrium forthe hinge a, we obtainv ai sin ai cos aiu ai cos ai sin ai =pxh/e(c)v ai sin ai u ai sin ai cos ai=pyh/e92 22 2from these two equations, the unknowns u and v can be readily calculated in each particular case. after this,substitution o

26、f u and v into expression (b) gives us the force si in any bar of the system. it is seen that for thisproblem, direct consideration of the deformation of the system results in a substantial simplificationof thesolution, especial if there are a large number of bars since, independently of that number

27、, 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 reactionsx1,x2,x3, as the redundant quantities, the theorem of least w

28、ork yields a system of equations(a),each ofwhich 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 eachspan of the continuous beam as a simple beam on

29、 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 adjacentspans must be equal, the known three-angle equations are obtained. such equations are much simpler thanesq.(a) becauseno one of them

30、 contains more than three unknowns.eadbcfig 10.2another 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 for

31、ces 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 greatlysimplified by considering the deformation of the structure. this deformation is completely defined by the angle of rota

32、tion a of the joint a produced by the applied loads. when the magnitude of this angle isfound。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 equ

33、ilibrium of the endmoments at joint a.it is not to be concluded from the foregoing discussion that, in the analysis of a statically indeterminatesystem, 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

34、 jointsthe method of deformations described above would become very cumbersome, whereas the method of forcesused is extremely simple.in dealing with highly statically indeterminate systems, we usually find that regardless of whether we usea method of forces or a method of deformations, it becomes ne

35、cessary to solve a large number of simultaneouslinear algebraic equations with as many unknowns without regard to any particular problem of structuralanalysis, let us now consider asystem of such equations:a11 a12x2a1nxnc110a21 a22x2a2nxnc2 .am1am2x2amnxncmtheoretically, such a system of linear alge

36、braic equations can always be solved, but the progress ofsolution 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 condensedforma

37、ijxj=ci（ 10.1a）or simplyax=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 orderm*n. when there is only one column or one row of el

38、ements 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 a

39、lgebrathods.before proceeding with this, however, the reader should understand that the use of matrix methods instructural analysis holds no particular magic, nor does it represent any great advantage over the methodsdiscussed in preceding chapters so long as numerical calculations are to be made by

40、 hand. its real advantagelies in the fact that it lends itself particularlywell to the use of the electronic digital computer and therebyopens the door to the analysis of structural problems that would otherwise be too involved and complex tocope with by desk-calculator techniques. in the limited sp

41、ace available here, we shall be unable to disclosethe full power of the matrix approach, but it is hoped that the simple examples to be discussed will give thereader enough familiaritywith the method to. enable him to study the literature on tile subject to betteradvantage.2.matrixanalysis of contin

42、uousframescontinuous 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 solvingsuch problems is to have recourse to the electronic digital computer, and for t

43、his purpose a matrix formulationof the problem is the most advantageous. to illustrate a matrix method for such problems, we shallconsider 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, o

44、n the other hand, it will be extensiveenough 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

45、 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 onlybending deformation. under these assumptions, the deformation of the frame under load will be completelydefined by a set of s

46、ix displacements: namely, the horizontal displacements a , and b of the two floors andthe angles of rotation 1, 2， 3, 4 of the four rigid joints.when these six displacements have beenfound, all end moments can be calculated from the slope-deflection equations , and the problem is solved.we therefore

47、 introduce the column vector11j = a, b, 1,2, 3, 4 (a)and selectthis set of displacements asthe unknowns of the problem.fig10.13as a preliminary step to the calculation of the displacements , we first consider the two simple problemsillustrated in fig. 10.14. in fig. 10.14a, we give to the end a of a

48、 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 find2 2rab=12k/l mab=6k /lrab=12k/lmab=6k /l (b)in

49、 fig. 10.14b, the end a of the same beam is given an angle of rotation without allowing any lateraldeflection of a or any movement at all of the b. then, again using the slope-deflection equations 9.6 we findrab=6k/lmab=4k rab=6k/lmab=2k (b )the coefficients appearing in front of and in fqs. (b) and

50、 (b) are seen to represent the reactions, or forces of constraint, at the ends of the beam when the displacementsandare each equal to unity. thesequantities are called the stiffness influence coefficientsfor the beam corresponding to each type of displacement. for convenience of easy reference, thes

51、e stiffness coefficients are recorded in matrix form under eachbeam in fig. 10.14. ,now, let us return to the frame in fig. 10.13, remove all applied loads, and lock all joints against bothtranslation and rotation.this done, we remove just one constraint corresponding to any one of the six degreesof

52、 freedom of the system, say constraint j, and make there a unit displacementj=1. this will result in somedeformation of the structure consistent with the remaining artificialconstraints, and we proceed to calculate the reaction corresponding to each of the six degrees of freedom. that is, we calcula

53、te the system of externalforces and couples necessary to hold the structure in the assumed configuration defined by j=1. in the generalized sense, wedenote such an external reaction at i by sij regardlessof whether it is a force or a couple. thus, we define the stiffness influence coefficient sij as

54、 the external reaction at i due to an imposed unit12displacement at j when all other displacements are held equal to zero. in our example there will be 36 ofthese stiffness influence coefficients, and we now set about their calculation, making use of the single-memberstiffness coefficients shown in

55、fig. 10.14.let us begin in fig. 10.15a with a unit displacementa=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 thestructure in this configuration will act as shown in the figure, and their magnit

56、udes will be as listed beside thestructure.inthesecalculations,weconsiderlineardisplacements and forces positiveto the right,negative to the left,and angular displacements and couples positive when clockwise, negative whencounterclockwise. consider, for example, the calculation of sba. from fig. 10.

57、14a, we see that the2reaction at the- bottom of each upper-story column in fig. 10.15a is 12k/l acting to the left and that there are2two such columns; hence, sba=-24k/l 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

58、 in fig. 10.15a is counterclockwise and ofmagnitude 6k/l and that there is only one column; hence, s4a = -6k/l.the reader should check tile othervalues 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