外文翻譯--結(jié)構(gòu)分析的矩陣方法

1、<p>　　畢業(yè)設(shè)計(論文)外文資料翻譯</p><p>　　學(xué)院（系）： 機(jī)械工程學(xué)院 </p><p>　　專 業(yè)： 機(jī)械工程及自動化 </p><p>　　姓 名： </p><p>　　學(xué) 號：

2、 </p><p>　　外文出處： Theory of structures </p><p>　　Publisher:McGraw Hill </p><p>　　附 件： 1.外文資料翻譯譯文；2.外文原文。 </p><p>　　附件1：外文資料翻譯譯文</p&g

3、t;<p><b>　　結(jié)構(gòu)分析的矩陣方法</b></p><p><b>　　力法和應(yīng)變方法</b></p><p>　　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，在分析拱門和框架結(jié)構(gòu)時，分析步驟如下。首先，所有的冗余的約束被對應(yīng)的冗余的力（或力矩）取代，這些力的大小可通過基于應(yīng)變能的最小勢能原理

4、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。</p><p>　　在連續(xù)梁和框架分析中，另一種不同的方法曾被使用。在這個情況下，我們首先計算了結(jié)點的旋轉(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。</p><p>　　我們用一個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

5、和PY，作用在的5根懸于剛性基礎(chǔ)的等截面桿交點A處。因為桿數(shù)量大于A點平衡方程的數(shù)目，很明顯這是一個靜不定問題。一般來說，如果絞點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，通過最小勢能原理獲得應(yīng)變能表達(dá)式，進(jìn)而獲得所需的方程：</p><p>　　эU/эX1=0 эU/эX2=0 ……

6、 (a)</p><p>　　其中每個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。</p><p>　　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)P引起的位移量很小，那么第i桿的拉長量

7、 △li=vSin ai–u cosai</p><p>　　桿中的對應(yīng)的軸力為 Si=EAi(vSin ai–u cosai)/li= EAi(vSinａｉ–u cosai) Ｓin ai/h (b) </p><p>　　再寫出鉸點A的兩個平衡方程， 得</p><p>　　v ∑Ai Sinai Cos ai-u ∑Ai Cos2ai

8、Sin ai =Pxh/E (c) </p><p>　　v ∑Ai Sinai-u ∑Ai Sinai Cos ai=Pyh/E</p><p>　　從這兩個方程中，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。對于這個問題，可以看出，直接考慮系統(tǒng)變形使得問題解決簡單化，

9、尤其在遇到很多根桿的時候，無需考慮桿的多少，我們只需解2個方程而已。</p><p>　　在類似的方法下，對連續(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)生的多余的對應(yīng)反力X1,X2,X3，……，用最少勢能原理導(dǎo)出方程組(a)，其中每個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。對這個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算出這

10、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，因為他們沒有一個包含有3個以上未知數(shù)。</p><p>　　另一個運用應(yīng)變方法使問題大為簡單的代表例子是圖10.2所示系統(tǒng)。4個兩端固定桿剛接于a點。忽略桿中軸力影響，這個系統(tǒng)有7個冗余的元素，為解決這個問題，用最少勢能原理得到7個方程。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

11、用下交點旋轉(zhuǎn)的角度θa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a的末端力矩方程的基礎(chǔ)上只需一個方程即可解出變形。</p><p>　　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，在一個含有1個冗余度和10個結(jié)點的簡單桁架中，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。</p><p>　　在處理高次靜不定系統(tǒng)時，我們通常發(fā)

12、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：</p><p>　　………………………………..（10.1）</p><p>　　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

13、簡為：</p><p>　　[aij][xj]=[ci] （10.1a）</p><p>　　或簡記 Ax=c （10.1b）</p><p>　　方括號表達(dá)式中的每個數(shù)組(

14、或記法)被稱為一個矩陣。數(shù)（或記法)本身被稱為元素，當(dāng)矩陣有m行和n列時，矩陣被稱為m*n型。當(dāng)僅僅在矩陣有一列或一行元素時，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。</p><p>　　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

15、的優(yōu)勢在于它引導(dǎo)去更好的利用了電子計算機(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)會他的優(yōu)點。</p><p>　　2 連續(xù)結(jié)構(gòu)的矩陣分析方法</p><p>　　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，以致于在分析時要處理分析許多未知數(shù)。解決這類問題的唯一的可行方法是求助于電

16、子數(shù)字計算機(jī)。并且為實現(xiàn)這個目的，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)時所有的步驟、過程。</p><p>　　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

17、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個位移量決定。即，兩個水平位移δa，δb四個交點處的旋轉(zhuǎn)角度θ1,θ2，θ3,θ4。6個位移量求出來以后，所有末端力矩可通過力位移方程計算出，這個問題就解決了。因此，我們介紹列向量</p><p>　　[δj]={δa，δb，θ1,θ2,θ3,θ4} (a) </p><p>　　并將

18、這一系列位移量作為問題未知量。</p><p><b>　　圖10.13</b></p><p>　　作為計算位移量的第一步，我們首先考慮圖10.14舉例說明了的2個簡單的問題。在圖10.14a中，在兩端固定的等截面梁AB的端點A作用一個位移δ，A沒有任何旋轉(zhuǎn)運動，B沒有任何移動。那么，A、B兩點的反力根據(jù)方程（9.6）很容易就計算出了。并且我們發(fā)現(xiàn)</p>

19、;<p>　　Rab=12kθ/l2 Mab=6kθ/l Rab=12kθ/l2 Mab=6kθ/l (b)</p><p>　　在圖10.14b中，相同梁的端點A只有一個旋轉(zhuǎn)角度θ，不允許A有任何側(cè)面移動，端點B也沒有任何移動。接著，再使用應(yīng)力--變形方程{9.6}，我們發(fā)現(xiàn)</p><p>　　Rab=6kθ/l

20、 Mab=4kθ Rab=6kθ/l Mab=2kθ (b’)</p><p><b>　　圖10.14</b></p><p>　　在方程(b)和(b')中，出現(xiàn)在δ和θ前面的系數(shù)代表梁端部的反力、力或約束，而此時位移δ和θ都是單位位移。對應(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

21、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。</p><p>　　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，并且并交點處無傳遞和旋轉(zhuǎn)。完了后，我們移開與系統(tǒng)6個自由度之一相對應(yīng)的任一約束，叫約束j，并給予單位位移δj=1。這將導(dǎo)致與這個人為約束相一致的結(jié)構(gòu)變形，接著我們計算出6個自由度對應(yīng)的其余結(jié)果。那就是說，在假定δj=1的情況下，計算出了支持結(jié)構(gòu)系統(tǒng)所需要的外力和外力偶。

22、總的來說，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。在這個例子中將有36個這些剛度影響系數(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)完成整個系統(tǒng)（剛度影響系數(shù)）的計算。</p><p>　　在圖10.15a中，在單位位移δa=1時，即最高的地板的側(cè)面的位移為一個單位移，所有的另外的位移均相等為零。

23、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)，角位移順時針方向為正，反時針方向為負(fù)。例如，Sba的計算，見圖10.14a，我們看到每個頂層列的底部的反力。圖10.15a左部有2個如此的列并且（結(jié)果）是12k/l2；因此，圖上標(biāo)注Sba=-24k/l2。再考慮S4a的計算.由圖10.14a的結(jié)果，圖10.15a，列的4a的反作用力矩是反時針方向的，其大小為6k/l并

24、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。</p><p>　　下一步，在圖10. 15b中，設(shè)單位位移δb= 1，即中間層的單元的一個單位水平位移，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θ1=1,θ2=1,θ3=1,θ4=1對應(yīng)的誘導(dǎo)外力被標(biāo)注在圖10.15c，d，e，f，這就完成了整個結(jié)構(gòu)的影響系數(shù)的計算。

25、</p><p>　　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?lt;/p><p>　　可以觀察到這是一個對稱矩陣，并且這種對稱來自于協(xié)調(diào)理論的</p><p>　　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算出任何數(shù)據(jù)組合位移δj的條件下支持框架結(jié)構(gòu)所需的外力。例如，要求外力是<

26、;/p><p>　　Fa= Saaδa+ Sabδb+ Sa1θ1+ Sa2θ2+ Sa3θ3+ Sa4θ4</p><p><b>　　要求外力偶是</b></p><p>　　M1 = S1aδa+ S1bδb+ S11θ1+ S12θ2+ S13θ3+ S14θ4</p><p>　　等等。然而，我們正在尋找那些在圖1

27、0.13系統(tǒng)所示的外力作用下的位移的一系列值，那些力是是實實在在的結(jié)構(gòu)負(fù)載。真實的位移集合已在系統(tǒng)的代數(shù)方程中定義了</p><p>　　其中符號相反的ql/12和-ql/12表示梁34的端部力矩，即結(jié)點3和4各自的不平衡力矩。介紹矩陣記法</p><p>　　{Fi}={Pa Pb 0 Qa ql2/12 -ql2/12} (d)

28、 它被稱為負(fù)載矩陣， 例 (I0.17)的矩陣記為</p><p>　　[Sij][ δj] = [Fi] (10.17a) </p><p>　　在這個方程出現(xiàn)的3個矩陣各自表達(dá)為(c), (a)和(d)。</

29、p><p>　　例(I0.17a)，方程位移的解為</p><p>　　[δi] = [Sij][Fi] </p><p>　　我們注意到求解需要剛度矩陣[Sij]的逆矩陣， 這時我們就需要計算機(jī)的幫助了。 </p><p>　　附件2：外文資料翻譯原文</p><p>　　Matrix methods in struct

30、ural analysis</p><p>　　FORCE AND DEFORMATION METHODS</p><p>　　The various methods of analysis of statically indeterminate systems that have been used in preceding chapters fall into two distinct

31、 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 magnitud

32、es 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 proc</p><p>　　In the analysis of continuous beams and frames, a

33、 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

34、of continuous beams represents again the kind of approach. Such procedure is called the method of deformation.</p><p>　　To illustrate, on the same example, the distinction between the two methods, let us con

35、sider 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

36、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</p><p>　　Each

37、 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.</p><p>　　To solve the same problem, Nervier

38、 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 hing

39、e 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</p><p>　　And the corresponding axial force in the bar becomes Si=EAi(v S

40、inai –u cosai)/li= EAi(v Sinai –u cosai) Sinai/h (b) Writing now the two equations of equilibrium for the hinge A, we obtain </p><p>　　v ∑Ai Sinai Cos ai—u ∑Ai Cos ai Sin ai =Pxh/E

41、 (c) </p><p>　　v ∑Ai Sinai—u ∑Ai Sinai Cos ai=Pyh/E</p><p>　　From these two equations, the unknowns u and v can be readily calculated in each particular case. After this, substitution of u

42、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, e

43、special if there are a large number of bars since, independently of that number, we have to solve only two equations.</p><p>　　In a similar way, direct consideration of the deformations simplifies the analys

44、is 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),

45、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 b</p><p>

46、;　　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

47、. 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 si

48、mplified by considering the deformation of the structure. This </p><p>　　It is not to be concluded from the foregoing discussion that, in the analysis of a statically indeterminate system, a method of deform

49、ations is always superior to a method of forces. For example, in the case of a simple truss having one redundant reaction and ten joints</p><p>　　The method of deformations described above would become very

50、cumbersome, whereas the method of forces used is extremely simple.</p><p>　　In dealing with highly statically indeterminate systems, we usually find that regardless of whether we use a method of forces or a

51、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

52、 system of such equations:</p><p>　　………………………………..</p><p>　　Theoretically, such a system of linear algebraic equations can always be solved, but the progress of solution becomes cumbersome as th

53、e 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</p>

54、<p>　　[aij][xj]=[ci] （10.1a）</p><p><b>　　Or simply</b></p><p>　　Ax=c （10.1b）</p><p>　　Each ar

55、ray 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 t

56、here 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

57、 the left-hand side of the system of equations above.</p><p>　　Before proceeding with this, however, the reader should understand that the use of matrix methods in structural analysis holds no particular mag

58、ic, 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 particula

59、rly well to the use of the electronic digital computer and thereby opens the door to the analysis of structural problems that would othe</p><p>　　2.MATRIX ANALYSIS OF CONTINUOUS FRAMES </p><p&

60、gt;　　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 su

61、ch 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

62、here a two-story building frame as shown in Fig.（10.l 3） On the </p><p>　　For simplicity, we assume that each member has the same length l and the same flexural rigidity EI so that the stiffness factors are

63、 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. Unde

64、r these assumptions, the deformation of the frame under load will be completely defined by a set of six displacements: namely, the horizontal displa</p><p>　　[δj] = {δa, δb, θ1, θ2, θ3, θ4} (a)

65、 </p><p>　　and select this set of displacements as the unknowns of the problem.</p><p><b>　　Fig10.13</b></p><p>　　As a preliminary step to the calculation of the displac

66、ements , 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 δ, without allowing any rotation of the tangent

67、 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</p><p>　　Rab=12kθ/l Mab=6kθ/l Rab=12kθ/l Mab=6k

68、θ/l 　　　　　　　　 (b)</p><p>　　In Fig. 10.14b, the end A of the same beam is given an angle of rotation θ without allowing any lateral deflection of A or any movement at all of the B. Then, again using the slop

69、e-deflection equations {9.6} we find</p><p>　　Rab=6kθ/l Mab=4kθ Rab=6kθ/l Mab=2kθ 　　　　　　　　　　 (b’)</p><p>　　The coefficients appearing in front of δ andθin Fqs. (b) and (b') are seen t

70、o represent the reactions, or forces of constraint, at the ends of the beam when the displacementsδ andθare each equal to unity. These quantities are called the stiffness influence coefficients for the beam correspondi

71、ng 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. ,,</p><p>　　Now, let us return to the frame in Fig

72、. 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

73、, and make there a unit displacementδj=1. This will result in some deformation of the structure consistent with the remaining artificial constraints, and we proceed to calculate the reaction corresponding to each of the

74、six degrees of freedom</p><p>　　Let us begin in Fig. 10.15a with a unit displacementδa=1, that is, a unit lateral displacement of the top floor, all other displacements being held equal to zero. Then, the e

75、xternal 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

76、 forces positive to the right, negative to the left, and angular displacements and couples positive when </p><p>　　Next, in Fig. 10. 15b, we make a unit displacement δb = 1, that is, a unit horizontal displa

77、cement 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 shown in tile figure. Deformation p

78、atterns corresponding toθ1=1, θ2=1, θ3=1, θ4=1 and the corresponding induced external forces are shown in Fig. 10.15c, d, e, f. This completes the calculation of the influence coefficien</p><p>　　We now as

79、semble these stiffness coefficients in the form of a square matrix, called the stiffness matrix for the structure. It is written in the order a, b, 1, 2, 3, 4, both as to rows and as to columns, and becomes</p>&

80、lt;p>　　It will be noted that this is a symmetric matrix and that such symmetry is to be expected from the reciprocal theorem.</p><p>　　Having the stiffness influence coefficients as shown in the matrix (c

81、) above, we may now use the principle of position to calculate the external forces required to hold the frame in any configuration defined by an arbitrary set of values of the displacements δj. For example, the required

82、external force at a would be</p><p>　　Fa = Saaδa+ Sabδb+ Sa1θ1+ Sa2θ2+ Sa3θ3+ Sa4θ4</p><p>　　The required external moment at ! would be</p><p>　　M1 = S1aδa+ S1bδb+ S11θ1+ S12θ2+ S13

83、θ3+ S14θ4etc. However, we are looking for a set of values of the displacements that will result from the particular system of external forces shown in Fig. 10.13, i.e., the actual loading on the structure. Thus the true

84、 set of displacements is defined by the system of algebraic equations</p><p>　　where ql/12 and -ql/12 represent, with reversed signs, the fixed-end moments for the girder 34, that is, the unbalanced moments

85、at the joints 3 and 4, respectively. Introducing the matrix notation</p><p>　　{Fi}={Pa Pb 0 Qa ql/12 - ql/12} (d) </p><p>　　which is called the load mat

86、rix, Eqs. (I0.17), in matrix notation, become </p><p>　　[Sij][δj] = [Fi] (10.17a) </p><p>　　The three matrices appearing in this equation are def

87、ined by expressions (e), (a), and (d), respectively. Having Fq. (I0.17a), the displacements are found from the equation</p><p>　　[δi] = [Sij]-1[Fi] </p><p>　　We note that this solution requires