弹性力学刘维龙

1、(9.5.1)72 =Where.The volume strain give n by Eq. (8.8.6) becomes090801109 刘维龙 090801117 刘帅 192194 页原文Thus , for an axisymmertric problem , if we succeed in finding a harm onic functionY(r , z) so that the displaceme nt comp onents (948) and the stress comp onents (9410) satisfy all the boundary cond

2、itions , the right solution is obtained.When a displacement potential exists, the volume strain e is constant throughout the body concerned , as shown by Eqs. (943) and (944). Since such circumstancesrarely occur , few problems can be thoroughly solved by using displacement potential only .He nee ,

3、for many practical problems, we have to resort to more complicated displaceme nt fun cti ons. Sometimes, however, a displaceme nt pote ntial can be used as a suppleme ntary displaceme nt function to reduce the mathematical operati ons.9.5 LOVE ' AND GALERKlNS DISPLACEMENTFUNCTIONSFor the solutio

4、n of axisymmertric spatial problems, A.E.H. Love introducedadisplacement function二(r,z) and expressed the displacement components as:1 8匕ur= 2G讥"箱2(1-肿-(9.5.2)With Eqs.(9.5.1) and 99.5.2), the frist of Eqs. (9.4.6) is ide ntically satisfied fun ctio n. Substitut ing Eqs.(9.5.1) and (9.5.2) into

5、 Eqs. 99.1.4) and noting thatand the second equation requireswhich shows that ? must be a biharmonic12G E ,we have°违何一那4 L弘(9.5.3)胡(i)宀詠biharm on ic fun ctio nThus, an axisymmertric problem can be solved if we succeed in finding a properso that the displacement components given by Eqs. (9.5.1)

6、and the stress comp onents give n by Eqs. (9.5.3) satisfy the boun dary con diti ons.For the solutions of spatial problems in general,B.Galerkin generalizedthe Love'sdisplaceme nt fun cti on and expressed the displaceme nt comp onents in terms of threedisplacement functions 二(x, y, z),(x ,y, z),

7、 一 (x, y, z) as1Li =2G2"违除玮Jv=2G2(冋峙粧矢翌|,诂*曲-器+娄切12G(9.5.1)a1%3/Wherey 止 .The volume strain give n by Eqs.(9.1.2) becomesSubstitutingEqs.(9.5.4) and (9.5.5)displaceme nt fun cti ons must satisfy(9.5.5)in toEqs.(9.4.1).we see that the three二十 is, theymust be biharm onic functions. Substituti on

8、of Eqs. (9.5.4) and (9.5.5) into Eqs. (9.1.1) yields込严2(1-勺+时%q 二 2(1-#)?V=+d.汀八*护(9.5.6)二（1一“）=（7尹卄整电(9.5.7)Thus, a spatial problem in general can be solved if we succeed in finding threebiharm on ic functionsso thatthe displacementcomponents given by Eqs.(9.5.4) and the stress comp onents give n b

9、y Eqs.(9.5.6) and (9.5.7) satisfy the boun dary con diti ons.9.6 CONCENTRATED NORMAL LOAD ON BOUNDARY OF A SEMI-INFINITE BODYLet us consider a semi-infinitebody under the action of a concentrated normal loadPon the boundary plane (Fig. 9.6.1). Evidently, this is an axisymmetric problemwith the line

10、of action of Pas the axis of symmetry. Hence, the point of application of P is take n as the origi n of coord in ates and the line of actio n of P as z axis.On the non-loaded part of the boun dary pla ne, we have the boun dary con diti ons丄 J、 -(9.6.1)= a.- - r(9.6.2)Fig. 961At the origin , there is

11、 a small part of the boundary plane subjected to surface forces. These surface forces have a resultantP but their distributionis not given.Now, let us con sider a horiz on tal pla ne sect ion at a dista nee z from the boun dary pla ne. The no rmal stresses on this secti on mustbe in equilibrium with

12、 the surfaceforces and, consequently, in equilibrium with the load P. Noting the condition of axial symmetry, we have the followingequilibrium equation in place of the boundarycon diti on:(2rV+P=0.(963)It is evident that the stresses and strains in the body vary with the distanee from the load and t

13、he volume stra in? would not be con sta nt. Hence, a right soluti onof the problem cannot be obtained by taking a displacement function. According to dimensionalanalysis,the expression for any stress component must be P divided byquadratic terms of the linear quantities r, z and R. It follows from E

14、qs. (9.5.3)that the displacement function 一 must be P multiplied by linearterms of r, z andR. Now we assume 二 to be the product of a constant A, and the biharmonicfunctionR:占 + 二(9.6.4)Substitution of Eq. (9.6.4) into Eqs. (9.5.1) and (9.5.3) yields翻译:这样，对于一个对称性问题，如果找到适当的调和函数（r,z）,使得公式（）给出的位移分量和（）式给

15、出的盈利分量能够满足边界条件，就得到该问题的正确解答。应当指出：并不是所有的问题中的位移都是有势的，因此，位移势函数并不是在所有的问题中都存在的。当人也就很明显，用位移函数去解决问题。并不一定能成功。实际上，如 果位移势函数存在，则又e=?2 Y=C表示体积应变在整个弹性体中是常量，这种情况当然是很特殊的，因而位移势函数所能解决角度的问题是很少的。下一节所介绍的位移函数，则可解决较多的问题，但是，为了减少运算。有时也可结合应用位移势函数。§9-5拉普位移函数及加辽金位移函数为了求解对称性为他，拉普引用一个位移函数（r,z）,把位移分量表示成为ur 二-2(9.5.1)2G.z2其中r

16、 ；：r1-将表达式（）代入，贝U式变为1-2e 二2E 爲)将()和（9.5.2）代入轴对称方程（）,可见位移函数匚应满足的条件是.4 ' =0这就是说,应当是重调和函数。将（9.51 ）和（）代入（）注意2G E，我们可以得到微分方程ar2亠C):z: r亠(心£)(9.5.3):zr :；: ：22(2-审22；z4zr于是可见，对于一个轴对称问题，只需找到恰当的重调和函数'（r,z）使得（）式给出的位移分量和（）式给出的应力能够满足边界条件，就得到该问题的正确解答。函数称为拉普位移函数，但有时也不恰当的呗称为拉普应变函数。为了求一般的、非对称的空间问题，伽辽金

17、吧位移函数加以推广。他医用三个位移函数（x, y, z）, （x, y, z）, （x, y,z）把位移分量表示成为1 |2u 2(1 - 卩 2()2G- x :x - y - z(9.5.4)"丄2(1 一八 2 -丄()2G- x ；x: y - z1 |2"乔页-e-)其中2 2 2将式（）代入式变为1-2" _.2 -： -： e 2(2Gx -y(9.5.5)将（9.5.4）和（）代入（9.4.1 ）可见上丝三个位移函数满足是条件是=0，4=0，4=0.这就是说，三个位移函数应当是重调和函数。将（9.5.4)和（）代入（）得应力分量表达式= 2(1)L

18、 2 .(!2= 2(1 _)： 2(! T2%=2(._ 2.(!盲2£z）（）ccter一 2）（ 一一一）:y: x: y： zd2、/兌亠州亠换、 二J爲(9.5.6)2Gz ；z y z-yz=d2 二2）丄二二）爲 zxW(22 )czCx:y：z -x -y :z2 )：*x : x ：-y : z(9.5.7)xy= (7( _22 )rxrx-x' y,，'使得（）位移对于一个一般的空间问题，只要找到三个恰当的重调和函数 分量和（）和（）应力分量满足边界条件得到该问题的正确解答§ 9-6半空间体在边界上受法向集中力设有半空间体，体积不计，在其边界法向集中力 卩，图（）这是一个对称 性问题二对称轴就是力P的作用线。因此把Z轴放在P的作用线上，坐标原点就 放在P的作用点('- Z)z =0,r 二0fl)应力边界要求为图961此外，还应有这样的边界条件：在0点附近的一校部分边界上，有一组面力作用，他的分布不明确，但已知他等效集中力P。在半空间体的任何一水平面上的应力，