《弹性力学》（双语版） 第八章 空间问题

Three. Physical equations For an isotropic body, the relations between deformation components and stress components are as follows E oruro Dr E= + X E = y E These are physical equations for spatial problems If stress components are denoted by strain components, physical equations can be written as: o =e+2GE T=G o,,=he+2G8,I=Gy o=e+ 2Ga I=G y where:e=8+8 +8 E +)-2山 11

11 Three. Physical Equations For an isotropic body, the relations between deformation components and stress components are as follows:  ( )  ( )  ( ) z z x y y y z x x x y z E E E                = − + = − + = − + 1 1 1 xy xy zx zx yz yz G G G       1 1 1 = = = These are physical equations for spatial problems. If stress components are denoted by strain components, physical equations can be written as: z z y y x x e G e G e G          2 2 2 = + = + = + xy xy zx zx yz yz G G G       = = = where： x y z e =  +  +  ( )( )   1+ 1− 2 = E

空间题 三物理方程 对于各向同性体,形变分量与应力分量之间的关系如下: E 川o,+σ )I r 6,=Bl,从(+o,)=a :-以x+o,yxG2x 这就是空间问题的物理方程。 将应力分量用应变分量表示,物理方程又可表示为: o= ne+2G8 T=Gy o,=he+2GEyT=Gyx o=n e+ 2Ga =Gr 其中:。e=E,+En+E Eu 久= 1+)(1-2) 12

12 三 物理方程 对于各向同性体，形变分量与应力分量之间的关系如下：  ( )  ( )  ( ) z z x y y y z x x x y z E E E                = − + = − + = − + 1 1 1 xy xy zx zx yz yz G G G       1 1 1 = = = 这就是空间问题的物理方程。 将应力分量用应变分量表示，物理方程又可表示为： z z y y x x e G e G e G          2 2 2 = + = + = + xy xy zx zx yz yz G G G       = = = 其中： x y z e =  +  +  ( )( )   1+ 1− 2 = E

Four Equations of compatibility Differentiate the second and the third formula of geometric equations at the left. Adding these two, we get ae, ac a 03w Substitute the fourth formula of geometric equations into the above equation, we get aE, a (a) ayaz Similarly ax (b) a y=0r dy Ox Oxy 13

13 Four Equations of Compatibility Differentiate the second and the third formula of geometric equations at the left.Adding these two,we get           +      =    +    =   +   y w z v z y y z w y z v z y y z 2 2 3 2 3 2 2 2   Substitute the fourth formula of geometric equations into the above equation, we get z y y z yz y z    =   +      2 2 2 2 （a） Similarly           =   +      =   +   y x x y x z z x x y xy z x z x       2 2 2 2 2 2 2 2 2 2 （b）

空间题 四相容方程 将几何方程第二式左边对z的二阶导数与第三式左边对 y的二阶导数相加,得 a-8,ae a Ow a Ov aw 022*0y2 0v02 020y 2 Ovo= a2 将几何方程第四式代入,得 a8 dr (a) 02 同理06:,O26 azax (b) 8 千 axa 14

14 四 相容方程 将几何方程第二式左边对z的二阶导数与第三式左边对 y的二阶导数相加，得           +      =    +    =   +   y w z v z y y z w y z v z y y z 2 2 3 2 3 2 2 2   将几何方程第四式代入，得 z y y z yz y z    =   +      2 2 2 2 （a） 同理           =   +      =   +   y x x y x z z x x y xy z x z x       2 2 2 2 2 2 2 2 2 2 （b）

Differentiate the late three formulas of geometric equations separately for X,Y, Z, we get 1 OyOx ozon yu 0zOy OxY oz Oxdz Oyo From the above equations, we get 0n)0(。02l + OxOx oy 0z Ox( Oyo (a/=2 2 0n)-,06 Oyo 15

15 Differentiate the late three formulas of geometric equations separately for X,Y,Z,we get y z u x z v z x y w z y u y z x v y x w x xy z x yz    +    =      +    =      +    =   2 2 2 2 2 2    From the above equations,we get x y z u y z y z u x x y z x x xy z x yz     =           =              =          +   +   −       2 2 2 2 2 2