英漢雙語(yǔ)彈性力學(xué)

1、1 23 5-4 The single-valued condition of stress and displacement in multiply connected region5-3 Complex-variable representation of boundary condition 5-2 Complex-variable representation of stress and displacement 5-1 Complex-variable representation of stress function 5-6 Problem of infinite plane in

2、cluding hole5-5 Situation of infinite multiply connected bodyChapter 5 Complex-variable Methods for Plane Elasticity4 5-4 5-4 多連通域內(nèi)應(yīng)力與位移的單值條件多連通域內(nèi)應(yīng)力與位移的單值條件5-3 5-3 邊界條件的復(fù)變函數(shù)表示邊界條件的復(fù)變函數(shù)表示5-2 5-2 應(yīng)力和位移的復(fù)變函數(shù)表示應(yīng)力和位移的復(fù)變函數(shù)表示5-1 5-1 應(yīng)力函數(shù)的復(fù)變函數(shù)表示應(yīng)力函數(shù)的復(fù)變函數(shù)表示5-6 5-6 含孔口的無(wú)限大板問題含孔口的無(wú)限大板問題5-5 5-5 無(wú)限大多連體的情形無(wú)限大多連體的

3、情形第五章第五章 平面問題的復(fù)變函數(shù)法平面問題的復(fù)變函數(shù)法5Chapter 5 Complex-variable Methods for Plane Elasticity When solving plane problems by Cartesian coordinates or polar coordinates, the boundary of object is straight line or circular arc .To other boundary, for example ellipse, hyperbola, non-concentric circles and so o

4、n, we need use different curvilinear coordinates. Applying complex-variable can predigest these problems.In this chapter, we just introduce the simple application of complex-variable in elasticity.6第五章第五章 平面問題的復(fù)變函數(shù)法平面問題的復(fù)變函數(shù)法 直角坐標(biāo)及極坐標(biāo)求解平面問題，所涉及的物直角坐標(biāo)及極坐標(biāo)求解平面問題，所涉及的物體邊界是直線或圓弧形。對(duì)于其他一些邊界，例如體邊界是直線或圓弧形。對(duì)

5、于其他一些邊界，例如橢圓形、雙曲形、非同心圓等就要用不同的曲線坐橢圓形、雙曲形、非同心圓等就要用不同的曲線坐標(biāo)。應(yīng)用復(fù)變函數(shù)可使該類問題得以簡(jiǎn)化。本章只標(biāo)。應(yīng)用復(fù)變函數(shù)可使該類問題得以簡(jiǎn)化。本章只限于介紹復(fù)變函數(shù)方法在彈性力學(xué)中的簡(jiǎn)單應(yīng)用。限于介紹復(fù)變函數(shù)方法在彈性力學(xué)中的簡(jiǎn)單應(yīng)用。7 5-1 Complex-variable representation of stress function In chapter 2,we have proved, in plane problems, there is a stress function that is biharmonic functio

6、n of position coordinates, if body force is constant, i.e.04i, 1i, 1yzxzyzxzNow introduce complex variable z= xiy and zxiy to replace real variable x and y. Noticing,8 5-1 應(yīng)力函數(shù)的復(fù)變函數(shù)表示 在第二章中已經(jīng)證明，在平面問題里，如果體力是常量，就一定存在一個(gè)應(yīng)力函數(shù)，它是位置坐標(biāo)的重調(diào)和函數(shù)，即04現(xiàn)在，引入復(fù)變數(shù)z= xiy和 zxiy以代替實(shí)變數(shù)x 和y。注意i, 1i, 1yzxzyzxz9 We find the

7、transformation are)i()(zzyzzyzzyzzxzzxzzxzyxzyx2i,2ifurthermore,222222)(,)(zzyzzx10 可以得到變換式)i()(zzyzzyzzyzzxzzxzzxzyxzyx2i,2i進(jìn)而222222)(,)(zzyzzx11zzyx2222224Pzz224LetSo we can transform the function 04)(0224azzas0)(2224PFor02 P12zzyx2222224Pzz224令于是可將方程式04)(0224azz變換成為0)(2224P由02 P13 It is obvious k

8、nown P is harmonic function which can be obtained by real part of analytical function. Suppose f(z) as analytical function and let) )()(21zfzfPPzz224For)()(2141412zfzfPzz)( 4)(zzf) )( )( (212zzzzLetyieldsthus14 可知，P是調(diào)和函數(shù)可由解析函數(shù)的實(shí)部得到。設(shè)f(z)為解析函數(shù)，可令) )()(21zfzfPPzz224由)()(2141412zfzfPzz)( 4)(zzf) )( )(

9、(212zzzz令得則15)()( )( (21zzzzzzThen integrating with respect to z yields)(d)()()(21zgzzzzzz)(d)(zzzLet)( )(zzi.e.)()()()(21zgzzzzzIntegrating the above equation with respect to yieldszthus16 )()( )( (21zzzzzz再對(duì)z積分，得到)(d)()()(21zgzzzzzz)(d)(zzz令)( )(zz即)()()()(21zgzzzzz將上式對(duì) 積分，得到z則17 Notice the biharm

10、onic function on the left side of the above equation is a real function. It is obvious that the four terms on the right side must be conjugate two and two. The first two terms is conjugate, and the next two terms should be also conjugate:)()(zzgLetwe obtain the famous gusa formula )()()()(21zzzzzzit

11、 can be also written as)()(Rezzz18 注意上式左邊的重調(diào)和函數(shù)是實(shí)函數(shù)，可見該式右邊的四項(xiàng)一定是兩兩共軛，前兩項(xiàng)已經(jīng)是共軛的，后兩項(xiàng)也應(yīng)是共軛的：)()(zzg令即得有名的古薩公式 )()()()(21zzzzzz也可以寫成)()(Rezzz19 So in plane problems when body force is constant, stress function can be represented by two analytical functions of complex variable z, (z) and (z), named K-M f

12、unction. So solving plane problems is just selecting K-M function properly and determining any constant in them according to boundary condition. 20 于是可見，在常量體力的平面問題中，應(yīng)力函數(shù)總可以用復(fù)變數(shù)z的兩個(gè)解析函 (z)和(z)來(lái)表示，稱為K-M 函數(shù)。而求解各個(gè)具體的平面問題，可歸結(jié)為適當(dāng)?shù)剡x擇這兩個(gè)解析函數(shù)，并根據(jù)邊界條件決定其中的任意常數(shù)。215-2 Complex-variable representation of stress a

13、nd displacementAccording to the relation between stress component and stress function,yxxyxyyx22222One Complex-variable representation of stress component22 5-2 5-2 應(yīng)力和位移的復(fù)變函數(shù)表示應(yīng)力和位移的復(fù)變函數(shù)表示根據(jù)應(yīng)力分量和應(yīng)力函數(shù)的關(guān)系yxxyxyyx22222一一 應(yīng)力分量的復(fù)變函數(shù)表示應(yīng)力分量的復(fù)變函數(shù)表示23 We find Complex-variable representation of stress compon

14、ent,zzxyyx242222) )()()()(21zzzzzzfor)( Re4 )( )( 2zzzyxyieldshence, for24 可得到應(yīng)力分量的復(fù)變函數(shù)表示zzxyyx242222) )()()()(21zzzzzz由)( Re4 )( )( 2zzzyx可得而由25 224i2i 22222i 22zyxyxyxxyxy)( )( 2i 2zzzxyxyyields)( )( 2i 2zzzxyxyor26 224i2i 22222i 22zyxyxyxxyxy)( )( 2i 2zzzxyxy可得)( )( 2i 2zzzxyxy或27 Only given (z)

15、and (z), we can divide the right side of above equation into imaginary part and real part, from imaginary part we get xy, from real part we get y-x.)( Re4 )( )( 2zzzyx)( )( 2i 2zzzxyxyandis complex-variable representation of stress component. Of course by building equations, x、y 、xy can be represent

16、ed by (z) and (z) respectively, but that will make equations become lengthiness and its not convenient to use.28 只要已知(z)及 (z)，就可以把上述公式右邊的虛部和實(shí)部分開，由虛部得出xy，由實(shí)部得出y-x。 )( Re4 )( )( 2zzzyx)( )( 2i 2zzzxyxy和就是應(yīng)力分量的復(fù)變函數(shù)表示。當(dāng)然也可以建立公式，把x、y 、xy三者分開用(z)和 (z)來(lái)表示，但那些公式將比較冗長(zhǎng)，用起來(lái)很不方便。29 Two Complex-variable represen

17、tation of displacement component Assuming plane stress problems, according to geometrical equation and physical equationyieldsyyxyxxuE)1 ()(xyxxyyvE)1 ()(xyyuxvE)()1 (22222)1 ( )()(2)1 ( )()( 2xzzxxzzxuE30 二二 位移分量的復(fù)變函數(shù)表示位移分量的復(fù)變函數(shù)表示 假定為平面應(yīng)力問題。由幾何方程及物理方程yyxyxxuE)1 ()(xyxxyyvE)1 ()(xyyuxvE)()1 (22222)1

18、 ( )()(2)1 ( )()( 2xzzxxzzxuE可得31)i()(zzyzzyzzyzzxzzxzzxfor)( )()(zzzzzand notice )()()()( )()( )()(zzzzzzzzzzzzx )()()()( )()( )()(izzzzzzzzzzzzysimilarlyyields32 )i()(zzyzzyzzyzzxzzxzzx由于)( )()(zzzzz并注意到 )()()()( )()( )()(zzzzzzzzzzzzx )()()()( )()( )()(izzzzzzzzzzzzy同理可得33Successive integration o

19、f the above two equations with respect to x and y, lead toWhere f1 and f2 are arbitrary functions. Substituting above equations into the following equation2222)1 ( )()(i2)1 ( )()( 2yzzxyzzyvE)()1 ( )()(i2)()1 ( )()( 221xfyzzEvyfxzzEuyxyuxvExy2)()1 (2342222)1 ( )()(i2)1 ( )()( 2yzzxyzzyvE將上兩式分別對(duì)x及y積分

20、，得)()1 ( )()(i2)()1 ( )()( 221xfyzzEvyfxzzEu其中的f1及f2為任意函數(shù)。將上式代入式y(tǒng)xyuxvExy2)()1 (235 )i()(zzyzzyzzyzzxzzxzzxfor )()( i)()(i )()(i )()(zzzzzzzzzzzzy )()( i)()(i )()(i )()(izzzzzzzzzzzzx36 )i()(zzyzzyzzyzzxzzxzzx由于 )()( i)()(i )()(i )()(zzzzzzzzzzzzy )()( i)()(i )()(i )()(izzzzzzzzzzzzx37 and yieldsxxf

21、yyfd)(dd)(d11Thus we can find the displacement of rigid body f1(y)u0y，f2(x)v 0 xWe get)(dd)(dd)1 ( 2)(dd)1 ( )()(i 2)(dd)1 ( )()(22122212xfxyfyyxxfxyxzzxuyfyyxzzyuxvyuE38 )(dd)(dd)1 ( 2)(dd)1 ( )()(i 2)(dd)1 ( )()(22122212xfxyfyyxxfxyxzzxuyfyyxzzyuxvyuE從而得到xxfyyfd)(dd)(d11于是得到剛體位移 f1(y)u0y，f2(x)v 0

22、x故有39 If neglecting displacement of rigid body ,we have)i)(1 ()(4)i(yxzvuEzyx2ifor )()()()(21zzzzzzyields)()()()( )()(2izzzzzzzzzyx40 若不計(jì)剛體位移，則有zyx2i由式 )()()()(21zzzzzz)()()()( )()(2izzzzzzzzzyx得到)i)(1 ()(4)i(yxzvuE41 let the result back substitution, and on the two side divide 1+ yields)()()(13)i(1

23、zzzzvuEThis is complex-variable representation of displacement component .If (z) and (z) are given, we can divide the real part and imaginary part of the right side of the above formula, and u and v can be solved. The above formula is educed in plane stress problem. To plane strain problem, we need

24、replace E with E/(1 2) and with /(1 )。42將結(jié)果回代,并兩邊除以1+ 得 這就是位移分量的復(fù)變函數(shù)表示。若已知(z)及 (z)，就可以將該式右邊的實(shí)部和虛部分開，從而得出u和v。)()()(13)i(1zzzzvuE 上述公式是針對(duì)平面應(yīng)力情況導(dǎo)出的。對(duì)于平面應(yīng)變情況，須將式中的E改換為E/(1 2)， 改換為 /(1 )。435-3 Complex-variable Representation Of Boundary Condition To evaluate of every crunode in boundary, we need apply bo

25、undary condition of stress, i.e.:YmlXmlyxyxyxyxxyxyyx22222,andSubstitution into the above equation, givesYxmyxlXyxmyl222222445-3 5-3 邊界條件的復(fù)變函數(shù)表示邊界條件的復(fù)變函數(shù)表示 為了求得邊界上各結(jié)點(diǎn)處的值，須要應(yīng)用應(yīng)力邊界條件，即： YmlXmlyxyxyxyxxyxyyx22222,而代入上式，即得： YxmyxlXyxmyl22222245 As the figure shownl=cos(N,x)=dy/ds, m=cos(N,y)=-dx/ds,So t

26、he above equation can be rewritten as:YxsxyxsyXyxsxysy222222ddddddddYxsXysdd,ddThus, yields46 由圖可見，l=cos(N,x)=dy/ds, m=cos(N,y)=-dx/ds,于是，前式可改寫為：YxsxyxsyXyxsxysy222222ddddddddYxsXysdd,dd由此得： 47 Suppose A is a fix point in the boundary, and B is a arbitrary point, so the composition of forces from A

27、to B can be obtained by integrating of the above equation with respect to s from A to B,BABABABAyxyxxysxyssYXPPiiididddii)i()(zzyzzyzzyzzxzzxzzxSubstituting this formula48 設(shè)A是邊界上的固定點(diǎn),B為任意一點(diǎn)，則從A到B邊界上的合力，可用上式從A點(diǎn)到B點(diǎn)對(duì)s積分得到：)i()(zzyzzyzzyzzxzzxzzx將式BABABABAyxyxxysxyssYXPPiiididddii49 into the above equat

28、ion, and rearrangement yieldsBAyxzzzzPP )()( )( iiAdding a complex constant into stress function, which doesnt influence the stress. So we can let A of stress function as zero, and then in the boundary gives )()( )( iiyxPP)()( )()i( iyxPPorThis is boundary condition of stress.50 代入，整理得：BAyxzzzzPP )(

29、)( )( ii把應(yīng)力函數(shù)加上一個(gè)復(fù)常數(shù)，并不影響應(yīng)力。因此，可把應(yīng)力函數(shù)A處的值設(shè)為零，于是對(duì)于邊界上的有 )()( )( iiyxPP)()( )()i( iyxPP或這就是應(yīng)力邊界條件。51 To boundary condition of displacement)()()(13)i(1zzzzvuEssvvuu,Substituting them into the following equationWe can obtain complex-variable representation of boundary condition of displacement in plane

30、stress problem.)i(1)()()(13ssvuEzzzz To plane strain problem, we need replace E with E/(1 2) and with /(1 )。52 對(duì)于位移邊界條件ssvvuu,將其代入下式即得平面應(yīng)力情況下位移邊界條件的復(fù)變函數(shù)表示)i(1)()()(13ssvuEzzzz)()()(13)i(1zzzzvuE 對(duì)于平面應(yīng)變，須將式中的E 改換為E/(1 2)，改換為 /(1 )。53 5-4 The single-valued condition of stress and displacement in multi

31、ply connected region When stress is determined, the stress function can still a arbitrary linear function, so the K-M function is not determined completely, so to the simply connected region, the K-M function can be determined by selecting the suitable coordinate. But to the multiply connected regio

32、n ,it is still a problem. In this section, the condition of the K-M function satisfied single-valued in multiply connected region is discussed. Suppose there is a multiply connected region that has a interior boundary C, and in the interior boundary C the external force vector is given. Generally mu

33、ltiform function is logarithmic function, we suppose 54 5-4 5-4 多連通域內(nèi)應(yīng)力與位移的單值條件多連通域內(nèi)應(yīng)力與位移的單值條件 應(yīng)力確定后，應(yīng)力函數(shù)仍可差一個(gè)任意的線性函數(shù)，這時(shí)K-M函數(shù)并未完全確定.對(duì)于單連通區(qū)域，可以通過選取適當(dāng)坐標(biāo)系等辦法，使得K-M函數(shù)完全確定;但對(duì)于多連通區(qū)域仍不能完全確定.本節(jié)討論K-M函數(shù)在多連通區(qū)域內(nèi)滿足單值的條件。 設(shè)有多連通區(qū)域，有一內(nèi)邊界C，設(shè)在邊界C上的外力矢量已給定。通常的多值函數(shù)是對(duì)數(shù)函數(shù)，我們?cè)O(shè)55 )()ln()()()ln()(zzzBzzzzAzfkkfkkDCWhere zk

34、is a arbitrary point in the interior boundary, f andf are single-valued analytical function(holomorphic function), and Ak and Bk are constants gives:56 )()ln()()()ln()(zzzBzzzzAzfkkfkkDC這里zk為內(nèi)部邊界內(nèi)的任意一點(diǎn)，f和f為單值的解析函數(shù)（全純函數(shù)），而Ak ，Bk為常數(shù)：57 kkkkkkBAi,iThe derivative of above function is single-valued, but

35、it is multivalued by itself, while z move around the circum once, the value of ln(zk) have a increment of 2i,so the increments of (z) and (z) is 2i Ak and 2iBk respectively, and according to the following formula ,the master vector of stresses )()( )( iizzzzYXBAthe master vector of stress(around the

36、 whole boundary) is on the left side ,and the increment is on the right side:ykxkkkPPBAi)(258 kkkkkkBAi,i前面的函數(shù)的導(dǎo)數(shù)是單值的，但他們本身是多值的，當(dāng)z繞周邊一周時(shí)，函數(shù)值ln(zk)產(chǎn)生一個(gè)增量2i, 于是(z)和 (z)的增量分別是2i Ak和2iBk,這時(shí)應(yīng)力主矢量按照公式)()( )( iizzzzYXBA左邊將得到應(yīng)力主矢量(沿整個(gè)邊界），右邊得到一增量：ykxkkkPPBAi)(259 While according to the following formula, dis

37、placementwill obtain increment, and according to single-valued the increment should be zero：ykxkkkPPBAi)(2andyields)1(2i)1(2iykxkkykxkkPPBkPPA)13(0kkBA)()()(13)i(1zzzzvuE60 這時(shí)位移按照公式)()()(13)i(1zzzzvuE也將得到增量，根據(jù)單值性這個(gè)增量應(yīng)為零：)13(0kkBAykxkkkPPBAi)(2結(jié)合可得到)1(2i)1(2iykxkkykxkkPPBkPPA61 )()ln()1 (2i)()()ln()1

38、 (2i)(ffzzzPPzzzzPPzkykxkkykxkthusWhile there is m interior boundary, let)()ln()1 (2i)()()ln()1 (2i)(f1f1zzzPPzzzzPPzmkkykxkmkkykxk62 )()ln()1 (2i)()()ln()1 (2i)(ffzzzPPzzzzPPzkykxkkykxk于是當(dāng)有m個(gè)內(nèi)邊界時(shí)，取)()ln()1 (2i)()()ln()1 (2i)(f1f1zzzPPzzzzPPzmkkykxkmkkykxk63 5-5 Situation of infinite multiply connec

39、ted body When exterior boundary of multiply connected body approach infinite farness, this multiply connected body become infinite multiply connected body, besides the above conditions, we need consider the ultimate situation of infinite farness. Regards origin of coordinate as the centre of circle,

40、 draw a big enough circle sR ,which include all interior boundary, To an arbitrary point in the elastomer, but beyond sR, giveszzzzzzzzzzzkkkkln21ln1lnln)ln(2the analytical function beyond sR64 5-5 5-5 無(wú)限大多連體的情形無(wú)限大多連體的情形 當(dāng)多連體的外邊界趨于無(wú)限遠(yuǎn)時(shí)，該多連體成為無(wú)限大的多連體，除上述條件外，還需考慮無(wú)限遠(yuǎn)的極限情況。 以坐標(biāo)原點(diǎn)為圓心，作充分大的圓周sR，將所有的內(nèi)邊界包圍在

41、其內(nèi)，對(duì)于sR之外，彈性體之內(nèi)的任意一點(diǎn)，可得到 zzzzzzzzzzzkkkkln21ln1lnln)ln(2在sR之外的解析函數(shù)65 )()ln()1 (2i)()()ln()1 (2i)(f1f1zzzPPzzzzPPzmkkykxkmkkykxkso)(ln)1 (2i)()(ln)1 (2i)(f*f*zzPPzzzPPzyxyxIt can be written asWhere Px,Py are the sum of surface forces in m boundary.66 )()ln()1 (2i)()()ln()1 (2i)(f1f1zzzPPzzzzPPzmkkykx

42、kmkkykxk于是)(ln)1 (2i)()(ln)1 (2i)(f*f*zzPPzzzPPzyxyx可寫為其中Px,Py為m個(gè)邊界上沿x,y方向的面力之和。67 Expand holomorphic function *f and*f in Multiply connected region by luolang series:nnnnzbzzaz)()(f*f*)(21)(12i1)(12i211nnnnyxyxyxzazanzPPzPPsoFor components of stresses in infinite farness is finite, the coefficients

43、 of n2 is zero.)2(0, 0naann68 將多連通區(qū)域內(nèi)的全純函數(shù)*f和*f展開為羅郎級(jí)數(shù)：nnnnzbzzaz)()(f*f*)(21)(12i1)(12i211nnnnyxyxyxzazanzPPzPP于是 由于在無(wú)窮遠(yuǎn)處的應(yīng)力分量應(yīng)該是有限的，級(jí)數(shù)中n2的系數(shù)應(yīng)為零。)2(0, 0naann69 )( )( 2i 2zzzxyxySimilarity, fromfor components of stress in infinite farness is finite, so)2(, 0nbn)()()()()()()()(0f11111f*0f1f*zzizbzizz

44、zizaziznnnnnnWhere neglecting the constant terms that have no relation to stresses.70 )( )( 2i 2zzzxyxy同樣從中，由于在無(wú)窮遠(yuǎn)處的應(yīng)力分量應(yīng)該是有限的，故有)2(, 0nbn)()()()()()()()(0f11111f*0f1f*zzizbzizzzizaziznnnnnn其中略去了和應(yīng)力無(wú)關(guān)的常數(shù)項(xiàng)。71 )(ln)1 (2i)()(ln)1 (2i)(f*f*zzPPzzzPPzyxyxso)()i()i()()()i()i()(0f11111f*0f1f*zzzbzzzzzazznn

45、nnnnWhere is no relation to stress calculate, it can be regarded as zero ,and10f10f)()(nnnnnnzbzzaz72 )(ln)1 (2i)()(ln)1 (2i)(f*f*zzPPzzzPPzyxyx于是)()i()i()()()i()i()(0f11111f*0f1f*zzzbzzzzzazznnnnnn其中與應(yīng)力計(jì)算無(wú)關(guān)，可取為零，而10f10f)()(nnnnnnzbzzaz73 1)(241)(12i1)(12i2nnnnnyxyxyxzazanzPPzPPAt this timeWhen z，fi

46、elds4yxSimilarity, when z, for)( )( 2i2zzzxyxyfields11i22i2xyxyFrom above equation, we can obtain the corresponding coefficient and we can also find in infinite farness the distributing of stress is symmetrical.74 1)(241)(12i1)(12i2nnnnnyxyxyxzazanzPPzPP這時(shí)當(dāng)z時(shí)，可得4yx同樣當(dāng)z時(shí)，由)( )( 2i2zzzxyxy可得11i22i2xyx

47、y從中可求得相應(yīng)的系數(shù)，并可以看到在無(wú)限遠(yuǎn)處，應(yīng)力的分布是均勻的。75 )(2)()(4)()(11xyxyyxcoefficients)()(i2)()(ln)1 (2i)()(4)()(ln)1 (2i)(0fxy0fzzzPPzzzzPPzxyyxyxyxthus76 )(2)()(4)()(11xyxyyx系數(shù))()(i2)()(ln)1 (2i)()(4)()(ln)1 (2i)(0fxy0fzzzPPzzzzPPzxyyxyxyx則775-6 Problem of infinite plane including hole Regards origin of coordinate

48、as the centre of circle, draw a big enough circle sR , which include all interior boundary, so to an arbitrary point in the elastomer but beyond sR, we have)(ln)1 (2i)()(ln)1 (2i)(f*f*zzPPzzzPPzyxyx785-6 5-6 含孔口的無(wú)限大板問題含孔口的無(wú)限大板問題 以坐標(biāo)原點(diǎn)為圓心，作充分大的圓周sR，將所有的內(nèi)邊界包圍在其內(nèi)，對(duì)于sR之外，彈性體之內(nèi)的任意一點(diǎn)，可得到)(ln)1 (2i)()(ln)1

49、 (2i)(f*f*zzPPzzzPPzyxyx79111f*1f*)i()()(nnnnnnzbzzzazz1)2(21) 1(111) 1() 1(1)1 (2i)( i1)1 (2i)( 1)1 (2i)( nnnyxnnnyxnnnyxznnazPPznzbzPPznzazPPz80111f*1f*)i()()(nnnnnnzbzzzazz1)2(21) 1(111) 1() 1(1)1 (2i)( i1)1 (2i)( 1)1 (2i)( nnnyxnnnyxnnnyxznnazPPznzbzPPznzazPPz81 1)1(00)()( )( kkkkkkkkkkzAzzBzzAz

50、rewritten as1100iiBAwhere) 2() 1() 1()1 (2i)1 (2i1111kbkBakAPPBPPAkkkkyxyx82 1)1(00)()( )( kkkkkkkkkkzAzzBzzAz改寫為1100iiBA其中) 2() 1() 1()1 (2i)1 (2i1111kbkBakAPPBPPAkkkkyxyx83To the points in the boundary of holeieaz 0i2i0i0ie)(ee)( e)( 111kkkkCkkkkCkkkkCkaAzzaAzaAz0i)2(2i112i00)2i(2ieeee)( e1kkkkkkk

51、kCaBaBBaBz)20 (84 對(duì)于孔邊上的點(diǎn)ieaz 0i2i0i0ie)(ee)( e)( 111kkkkCkkkkCkkkkCkaAzzaAzaAz0i)2(2i112i00)2i(2ieeee)( e1kkkkkkkkCaBaBBaBz)20 (85Substituting the above equations into the following equation2ie)( )( )( )( iWe can obtain the stress boundary condition of series forms of circle boundary in polar coord

52、inate. Assuming the external force is known and spread it by Fourier series,20iide)i(21eikkkkkppCCpp86 將上列各式代入2ie)( )( )( )( i就得到極坐標(biāo)下圓周邊界上的級(jí)數(shù)形式的應(yīng)力邊界條件。 設(shè)周邊上的外力為已知，并將其展開為傅氏級(jí)數(shù)20iide)i(21eikkkkkppCCpp87 2ie)( )( )( )( eikkkC0i -) 2(2i112i00i0i0iieeeeeeekkkkkkkkkkkkkkkkkkkkaBaBBkaAaAaACBy comparing coef

53、ficient of eik and e-ik, fields121102200CaBaACaBAA88 2ie)( )( )( )( eikkkC0i -) 2(2i112i00i0i0iieeeeeeekkkkkkkkkkkkkkkkkkkkaBaBBkaAaAaAC比較兩邊eik和e-ik的系數(shù)，可得121102200CaBaACaBAA89 1)1 () 3(222022kCaBaAkkCaACBaAkkkkkkkk1100iiBAFor the stresses condition in infinite farness, yields)(,2)()(,4)()(11xyxyyx90

54、 1)1 () 3(222022kCaBaAkkCaACBaAkkkkkkkk1100iiBA由無(wú)限遠(yuǎn)處的應(yīng)力條件，可得)(,2)()(,4)()(11xyxyyx91For the single-valued condition of displacement yields011 BA1211CaBaAandWe find1,11111CaBCaAforkkkkkkkkCaBaAkkCaACBaA222022)1 () 3(92由位移的單值條件有011 BA1211CaBaA及可求得1,11111CaBCaA再由kkkkkkkkCaBaAkkCaACBaA222022)1 () 3(93 )

55、3() 1()3()2(),(22200222022kCaAakBkCaACAaBCBaAkkkkkkkyieldsBy now, all coefficients have been solved.For example, assuming the uniform pressure around hole is p and the stress of infinite farness is zero.94 )3() 1()3()2(),(22200222022kCaAakBkCaACAaBCBaAkkkkkkk可求得至此，全部系數(shù)均已求出。例例 設(shè)孔周邊為均勻壓力p，無(wú)限遠(yuǎn)處的應(yīng)力為零。95

56、)0(0,i00,000kCpCpppBAppkThus) 3(,0022211kBApaBABAkkSo we get96 )0(0,i00,000kCpCpppBAppk則有) 3(,0022211kBApaBABAkk于是可求得97 zpazz2)( 0)( 0,20,22222uGpaupapaFinally, we getAccording to the above means, the general problems of infinite plane including hole can be solved.98 zpazz2)( 0)( 0,20,22222uGpaupapa

57、最后得到根據(jù)上述方法，圓孔口無(wú)限大板的一般問題都可以得到解決。99Exercise 5.1 Try to check-up the following Complex-variable (1)(2) zqzzqz2,4 iqzzz, 0Solution: The fundamental formula gives azxyRe4(1) Substituting zqzzqz2,4into (a)、(b) bizzzxyxy22100練習(xí) 5.1 試考察下列復(fù)變函數(shù)所解決的問題(1)(2) zqzzqz2,4 iqzzz, 0解: 基本公式為 azxyRe4 bizzzxyxy22(1) 將 z

58、qzzqz2,4分別代入(a)、(b)式101yieldsxyqxyxyiq2associate with the above two equations,yields0, 0,xyyxqThe given function can solve the problems that the rectangular sheet is under uniform pulling force q in axis x. As the figure 5.1(a) shown.(2) Substituting iqzzz, 0into (a) and (b), yieldsxyxyiiq22xyqqFigu

59、re 5.1(a)yx0102得xyqxyxyiq2聯(lián)立求解以上兩式,得0, 0,xyyxq 所給的函數(shù)可以解決矩形薄板在x方向受均布拉力q的問題.如圖5.1(a)所示(2) 將 iqzzz, 0代入(a),(b)兩式,得xyxyiiq22xyqq圖5.1(a)yx0103associate with the above two equations,yieldsqxyyx, 0, 0The given function can solve the pure shear problem of rectangular sheet. As the figure 5.1(b) shown.qqxy圖5

60、.1(b)Exercise 5.2 As the figure shown. Try to prove we can use complex-variable to solve the pure bending problem of beam of rectangular cross-section. 2288zIiMzzIiMzWhere I is the inertia moment of cross section of beam, M is the moment of flexion.MyxzySolution:The fundamental formula gives104聯(lián)立求解以