英漢雙語彈性力學(xué)11

1、1 2 3 Summarize 1-1 Movement differential equations of elastic objects 11-2 Without rotating wave and equal volume wave 11-3 Transverse wave and vertical wave 11-4 Spherical wave Chapter 11 Elastic Wave 4 概述概述 1-1 1-1 彈性體的運動微分方程彈性體的運動微分方程 11-2 11-2 無旋波與等容波無旋波與等容波 11-3 11-3 橫波與縱波橫波與縱波 11-4 11-4 球面波球面

2、波 第十一章第十一章 彈性波彈性波 5 Summarize: When elastic object bears loads in static force equilibrium conditions, not all the parts of object has displacement, distortion and stress. At the beginning of the loads, the parts which are more far from the loads have no impacts . After then ,the displacement , dist

3、ortion and stress caused by loads transmit to other places in a finite speed of wave. This wave is called elastic wave. This chapter will first give movement differential equations of elastic objects, then introduce some conceptions of elastic wave and simplify the equations according to different e

4、lastic waves, at last give the speed transmitting formulas of wave in infinite elastic objects. 6 概述概述 當(dāng)靜力平衡狀態(tài)下的彈性體受到荷載作用時，并不是在 彈性體的所有各部分都立即引起位移、形變和應(yīng)力。在作用 開始時，距荷載作用處較遠的部分仍保持不受干擾。在作用 開始后，荷載所引起的位移、形變和應(yīng)力，就以波動的形式 用有限大的速度向別處傳播。這種波動就稱為彈性波。 本章將首先給出描述彈性體運動的基本微分方程，然后 介紹彈性波的幾個概念，針對不同的彈性波，對運動微分方 程進行簡化，最后給出波在無限

5、大彈性體中傳播速度公式。 7 11-1 Movement differential equations of elastic objects The two assumptions are equal to the basic assumptions when we discuss static force questions. So the physic and geometry equations and elastic equations where stress component is expressed by displacement component , still are t

6、he same with movement equations at any instantaneous time. The only difference is that the equilibrium differential equations of static questions must be substituted by movement differential equations . This chapter we still adopt the assumptions: （1） Elastic objects are ideal elastic objects. （2） T

7、he displacement and distortion are tinny. 8 11-1 11-1 彈性體的運動微分方程彈性體的運動微分方程 上述兩條假設(shè)，完全等同于討論靜力問題的基本假 設(shè)。因此，在靜力問題中給出的物理方程和幾何方程， 以及把應(yīng)力分量用位移分量表示的彈性方程，仍然適用 于討論動力問題的任一瞬時，所不同的僅僅在于，靜力 問題中的平衡微分方程必須用運動微分方程來代替。 本章仍然采用如下假設(shè)： （1） 彈性體為理想彈性體。 （2） 假定位移和形變都是微小的。 9 Toward any tiny object , when we apply dAlembert theory

8、, we must consider stress , body force and the inertia force of elastic objects caused by acceleration . In space right-angle coordinate system, the x, y, z directions component of inertia force of every unite volume are: Where is the density of elastic objects. 2 2 t u 2 2 t 2 2 t w 10 對于任取的微元體，運用達

9、朗伯爾原理，除了 考慮應(yīng)力和體力以外，還須考慮彈性體由于具有加 速度而產(chǎn)生的慣性力。每單位體積上的慣性力在空 間直角坐標(biāo)系的x,y,z方向的分量分別為: 其中為彈性體的密度。 2 2 t u 2 2 t 2 2 t w 11 Because of the equilibrium relations ,we simplify them and get: The above formulas are called movement differential equations of elastic objects. They and geometry equations and physic eq

10、uations are the basic equations of movement questions of elasticity mechanics 0 2 2 t u X zyx zx yx x 0 2 2 t Y xzy xyzyy 0 2 2 t w Z yxz yz xzz 12 由平衡關(guān)系，并簡化后得： 上式稱為彈性體的運動微分方程。它同幾何方程和物理方程 一起構(gòu)成彈性力學(xué)動力問題的基本方程。 0 2 2 t u X zyx zx yx x 0 2 2 t Y xzy xyzyy 0 2 2 t w Z yxz yz xzz 13 Note 1: geometry equati

11、ons: x u x y y z w z zy w yz x w z u zx y u x xy 14 注注1 1：幾何方程：幾何方程 x u x y y z w z zy w yz x w z u zx y u x xy 15 Note2: physic equations )( 1 zyxx E )( 1 xzyy E )( 1 yxzz E yzyz E )1 (2 zxzx E )1 (2 xyxy E )1 (2 16 注注2 2：物理方程：物理方程 )( 1 zyxx E )( 1 xzyy E )( 1 yxzz E yzyz E )1 (2 zxzx E )1 (2 xyxy

12、E )1 (2 17 Because the displacement component is difficult to be expressed by stress and its derivative, so movement equations of elasticity mechanics are usually solved according to the displacement. Substitute the elasticity equations where stress components are expressed by displacement component

13、 into movement differential equations, and we let: Then we get: z w yx u e 0) 21 1 ( )1 (2 2 2 2 t u Xu x eE 0) 21 1 ( )1 (2 2 2 2 t Y y eE 0) 21 1 ( )1 (2 2 2 2 t w Zw z eE 18 由于位移分量很難用應(yīng)力及其導(dǎo)數(shù)來表示，所以彈 性力學(xué)動力問題通常要按位移求解。將應(yīng)力分量用位移 分量表示的彈性方程代入運動微分方程，并令： 得： z w yx u e 0) 21 1 ( )1 (2 2 2 2 t u Xu x eE 0) 21

14、 1 ( )1 (2 2 2 2 t Y y eE 0) 21 1 ( )1 (2 2 2 2 t w Zw z eE 19 These are the basic differential equations of movement equations solved by displacement. They are also called Lame equations. We need boundary conditions to solve Lame equations. Besides these we still need original conditions , because

15、displacement components are the function of time variable . In order to simplify calculation , usually we neglect body force. Now the movement differential equations of elastic objects can be simplified as : ) 21 1 ( )1 (2 2 2 2 u x eE t u ) 21 1 ( )1 (2 2 2 2 y eE t ) 21 1 ( )1 (2 2 2 2 w z eE t 20

16、 這就是按位移求解動力問題的基本微分方程，也稱 為拉密（Lame)方程。 要求解拉密方程，顯然需要邊界條件。除此之外， 由于位移分量還是時間變量的函數(shù)，因此求解動力問題 還要給出初始條件。 為求解上的簡便，通常不計體力，此時彈性體的運 動微分方程簡化為： ) 21 1 ( )1 (2 2 2 2 u x eE t u ) 21 1 ( )1 (2 2 2 2 y eE t ) 21 1 ( )1 (2 2 2 2 w z eE t 21 11-2 Without rotating Wave and equal volume wave 1. Without rotating waves With

17、out rotating wave means that in elastic objects , the distortion caused by waves is not rotating . That means rotating values of three vertical coordinates at any point in the elastic object are zero. x u y z w Where is potential function of displacement. This displacement is called without rotating

18、 displacement, and the elastic wave corresponds to the displacement are called without rotating wave. ),(tzyx Suppose the displacement of elastic objects can be expressed by : u ,v , w 22 11-2 11-2 無旋波與等容波無旋波與等容波 一、無旋波一、無旋波 所謂無旋波是指在彈性體中，波動所產(chǎn)生的變形不存在旋 轉(zhuǎn)，即彈性體在任一點對三個垂直坐標(biāo)軸的旋轉(zhuǎn)量皆為零。 假定彈性體的位移u,v,w可以表示成為： x

19、u y z w 其中 是位移的勢函數(shù)。這種位移稱為無旋位 移。而相應(yīng)于這種位移狀態(tài)的彈性波就稱無旋波。 ),(tzyx 23 proving：at any point of elastic object ,the rotating value of z axis is : So the rotating values of the three coordinates at any point of the elastic object are zero. y u x z substitute into the formula , we can get: x u y 0 z Similarly:

20、 0 x 0 y 24 證：在彈性體的任一點處，該點對z 軸的旋轉(zhuǎn)量 即彈性體的任一點對三個坐標(biāo)的旋轉(zhuǎn)量都等于零。 y u x z 將 代入，可得： x u y 0 z 同理 0 x 0 y 25 In the condition of without rotating displacement: 2 z w yx u e sou xxx e 222 similarly: 2 y e w z e 2 Substitute the above three formulas into movement differential equations without including body fo

21、rce, simplify it we can get wave movement equations of without rotating wave. 26 在無旋位移狀態(tài)下 2 z w yx u e 從而u xxx e 222 同理 2 y e w z e 2 將上三式代入不計體力的運動微分方程，并簡化后 得無旋波的波動方程 27 uc t u 22 1 2 2 22 1 2 2 c t wc t w 22 1 2 2 )21)(1 ( )1 ( 1 E cWhere: 1 c is the transmitting speed of without rotating wave in i

22、nfinite elastic objects. 28 uc t u 22 1 2 2 22 1 2 2 c t wc t w 22 1 2 2 )21)(1 ( )1 ( 1 E c 其中 1 c 就是無旋波在無限大彈性體中的傳播速度 29 Equal volume wave means that in the distortion caused by waves in elastic objects, the volume strain is zero. That means the volume of elastic object remains unchanged. 2. Equal

23、volume wave Suppose the displacement u, v, w of elastic objects satisfy the condition that the volume strain is zero: 0 z w yx u e This displacement is called equal volume displacement and the elastic wave corresponds to this displacement is called equal volume wave. 30 所謂等容波是指在彈性體內(nèi)，波動所產(chǎn)生的變形中體積應(yīng) 變?yōu)榱?/p>

24、 。即彈性體中任一部分的容積（即體積）保持不變。 二、等容波二、等容波 假定彈性體的位移u,v,w滿足體積應(yīng)變?yōu)榱愕臈l件，即： 0 z w yx u e 這種位移稱為等容位移。而相應(yīng)于這種位移狀態(tài)的彈性 波就是等容波。 31 Because ，so simplify movement differential equations without including body force we can get wave movement equations of equal volume wave: 0e uc t u 22 2 2 2 22 2 2 2 c t wc t w 22 2 2 2

25、where )1 ( 2 2 E c is the transmitting speed of equal volume waves in infinite elastic objects. 2 c 32 由于 ，故不計體力的運動微分方程，簡化后得等 容波的波動方程： 0e uc t u 22 2 2 2 22 2 2 2 c t wc t w 22 2 2 2 其中 )1 ( 2 2 E c 就是等容波在無限大彈性體中的傳播速度。 2 c 33 According to without rotating wave and equal volume wave , we give the con

26、clusion without proving that :in elastic objects , stress , strain and speed of particle transmit in the same way and the same speed as displacement. 34 對于無旋波和等容波，我們不加證明地給出如下結(jié)論： 在彈性體中，形變、應(yīng)力以及質(zhì)點速度，都將和位移以相 同的方式與速度進行傳播。 35 1. vertical wave definition The particle movement direction of elastic objects i

27、s parallel to the transmitting direction of elastic wave. （seen in the Fig.） 11-3 Vertical wave and transverse wave 36 一、縱波 定義 彈性體的質(zhì)點運動方向平行彈性波的傳播方向（圖示） 11-3 縱波與橫波縱波與橫波 37 Let x axis be the transmitting direction of waves, then the displacement components of any point in elastic objects are: ),(txuu

28、00w So: x u e And: 2 2 x u x e 0 y e 0 z e 2 2 2 x u u 0 2 0 2 w 38 將x軸取為波的傳播方向，則彈性體內(nèi)任取一點的位移 分量都有： ),(txuu 00w 從而 x u e 而 2 2 x u x e 0 y e 0 z e 2 2 2 x u u 0 2 0 2 w 39 Substitute them into movement differential equations without including body force, we can find that the second and third formulas

29、 are identical equations. So the first formula can be simplified as: 2 2 2 1 2 2 x u c t u where )21)(1 ( )1 ( 1 E c is transmitting speed of vertical wave in elastic objects. 1 c It is obvious that transmitting speed of vertical waves is the same as without rotating waves . In fact vertical wave is

30、 a kind of without rotating wave. 40 代入不計體力的運動微分方程，可見其第二、第三式成為 恒等式，而第一式簡化為： 2 2 2 1 2 2 x u c t u 其中 )21)(1 ( )1 ( 1 E c 為縱波在彈性體中的傳播速度。 1 c 顯然縱波的傳播速度與無旋波相同。事實上，縱波就是 一種無旋波。 41 The general solution of wave movement equations of vertical wave is:)()(),( 1211 tcxftcxftxu The physic meaning of the genera

31、l solution is: let us take the first item for example, function at a fixed time is function of x, and can be expressed by curve abc in Fig.(a) (suppose the form is like this ). After , the function becomes: )( 11 t cxf t )( 111 tctcxf If let ，then function becomes . Its format is similar to the orig

32、inal function . From the Fig we can see the only difference is transverse coordinate moves in level . So expresses the wave whose speed is along x positive direction . tcxx 11 )( 111 tcxf )( 11 tcxf tc 1 )( 11 tcxf 1 c 42 縱波波動方程的通解是： )()(),( 1211 tcxftcxftxu 該通解的物理意義：以其第一項為例，函數(shù) 在某 一個固定時刻將是x的函數(shù)，可以用圖(

33、a)中的曲線abc表示 （假設(shè)是這種形狀），在 時間之后，函數(shù)變?yōu)椋?)( 11 t cxf t )( 111 tctcxf 如果令 ，則函數(shù)可寫為 ，其形式 同原函數(shù) 完全類同，只是橫坐標(biāo)發(fā)生平移 tcxx 11 )( 111 tcxf )( 11 tcxftc 1 見圖。因此 表示以速度 向x軸正向傳播的波。)( 11 tcxf 1 c 43 Similarly expresses the wave whose speed is along x negative direction . The general solution expresses two waves transmittin

34、g in opposite directions (in Fig.b). And the speed is the modulus of wave movement equations. 1 c )( 12 tcxf 1 c 1 f c a b x (a) (b) tc 1 tc 1 tc 1 44 同理 ，表示以同樣速度 向x軸負向傳播的波。 整個通解表示朝相反兩個方向傳播的兩個波（如圖b），其 傳播速度為波動方程的系數(shù) 。 1 c)( 12 tcxf 1 c 1 f c a b x (a) (b) tc 1 tc 1 tc 1 45 2. Transverse wave definitio

35、n The partial movement direction of elastic object is vertical to the transmitting direction of elastic wave. Transmitting format of transverse wave 46 二、橫波二、橫波 定義 彈性體的質(zhì)點運動方向垂直于彈性波的傳播方向。 橫波的傳播形式 47 Still we let x axis be the transmitting direction of wave and y axis be displacement direction of part

36、ial. Then the displacement components of any point in the elastic object have: 0u ),(tx0w so0e and0 2 u 2 2 2 x 0 2 w Substitute them into movement differential equations without including body force, we can find that the first and third formulas are identical equations. So the second formula can be

37、 simplified as: 2 2 2 2 2 2 x c t )1 (2 2 E c 2 c is transmitting speed of transverse wave in elastic object. Because volume strain of transverse wave 48 仍然將x軸放在波的傳播方向，y軸為質(zhì)點位移方向，則 彈性體內(nèi)任取一點的位移分量都有 0u ),(tx0w 從而0e 而0 2 u 2 2 2 x 0 2 w 代入不計體力的運動微分方程，可見其第一、第三式成為恒 等式，第二式簡化為： 2 2 2 2 2 2 x c t )1 (2 2 E c

38、 2 c 為橫波在彈性體中的傳播速度。由于橫波的體積應(yīng)變 49 General solution to wave movement equations of transverse wave is: ，so transverse wave is equal volume wave 0e )()(),( 2221 tcxftcxftx It is obvious that the general solution expresses two waves transmitting in opposite directions . Its displacement is along y directi

39、on and the transmitting direction is along x direction . The transmitting speed is a constant . 2 c 50 橫波的波動方程的通解為： ，故橫波為等容波。0e )()(),( 2221 tcxftcxftx 顯然，整個通解表示朝相反兩個方向傳播的兩個波，它的 位移沿著y方向，而傳播方向是沿著x方向，傳播速度等于常 量 。 2 c 51 11-4 Spherical wave If elastic objects have spherical hole or spherical out surface

40、 ,when the spherical hole or spherical out surface bears spherical force , the elastic wave transmitting from the hole to the outside or form the surface to the inside is called spherical wave. Spherical wave is symmetrical of sphere. Apply basic differential equations of sphere symmetry, we get: 0)

41、 22 ( )21)(1 ( )1 ( 222 2 rr rr ku rdr du rdr udE Now . If we ignore body force , and use radial inertia force to substitute ),(truu rr 2 2 t u r r k 52 11-411-4 球面波球面波 如果彈性體具有圓球形的孔洞或具有圓球形的外表面， 則在圓球形孔洞或圓球形外表面上受到球?qū)ΨQ的動力作用時 ，由孔洞向外傳播或由外表面向內(nèi)傳播的彈性波，稱為球面 波。 球面波是球?qū)ΨQ的。利用球?qū)ΨQ的基本微分方程： 0) 22 ( )21)(1 ( )1 ( 222

42、2 rr rr ku rdr du rdr udE 此時， ，而不計體力時，用徑向慣性力),(truu rr 2 2 t u r 代替 ， r k 53 Then the above formula can be simplified as: We get： 0) 22 ( )21)(1 ( )1 ( 2 2 22 2 t u r u r u rr uE rrrr let： )21)(1 ( )1 ( 1 E c Suppose r ur then is potential function of displacement. Substitute it into formula (a), we

43、 get: ),(tr 0 122 2 2 2 1 2 2 t u cr u r u rr u r r rrr （a） 54 則上式簡寫成 即得： 0) 22 ( )21)(1 ( )1 ( 2 2 22 2 t u r u r u rr uE rrrr 令： )21)(1 ( )1 ( 1 E c 假定 r ur 則 是位移的勢函數(shù)。代入（a）式得 ),(tr 0 122 2 2 2 1 2 2 t u cr u r u rr u r r rrr （a） 55 So formula (b) can be written as Because rrrrr r rrr 221 2 2 3 3 2 2 2 2 2 2 trrt 0