《弹性力学》（双语版） 第十二章 薄板弯曲

Express stress components with bending we have w Eaw t u ax E(o21 ax E Ixy1+H 2 u oxo] These formulas indicate: the main stress components Ox,Oy,txy have linear distributing along board thickness (3) Differential Equation for Stretch Flexural Plane Ignoring body force, from the first two formulas of equilibrium equations, we get 21

21 （3）Differential Equation for Stretch Flexural Plane Ignoring body force, from the first two formulas of equilibrium equations, we get: These formulas indicate: the main stress components  x , y , xy have linear distributing along board thickness. z x y E w z x w y E w z y w x E w xy y x    + = −           +   − = −           +   − = − 2 2 2 2 2 2 2 2 2 2 2 1 1 1         Express stress components with bending ,we have w

薄城曲 将应力分量用挠度W表示,得: E 2+H ax E(o21 ax E Ixy1+H 2 u oxo] 上式说明,主要的应力分量σx沿板的厚度线性 分布。 (3)弹性曲面微分方程 在不计体力的情况下,由平衡方程的前二式得: 22

22 （3）弹性曲面微分方程 在不计体力的情况下，由平衡方程的前二式得： 上式说明，主要的应力分量 沿板的厚度线性 分布。 x y xy  , , 将应力分量用挠度 表示，得： z x y E w z x w y E w z y w x E w xy y x    + = −           +   − = −           +   − = − 2 2 2 2 2 2 2 2 2 2 2 1 1 1         w

aT 2 ax Substitute physical functions which stress components are denoted by bendin ng w into the ne above formulas and cancel terms we get ℃=1-2Ox Ez O V-M az aT ez a Vw 丿2 Because the bending w doesnt change along z axis, and the boundary conditions on top surface and under surface of sheet are (x)2=0,( 2) =0 23

23 z y x z x y zy y xy zx x yx   −   = −     −   = −         Substitute physical functions which stress components are denoted by bending into the above formulas and cancel terms,we get w w y Ez z w x Ez z z y z x 2 2 2 2 1 1    − =      − =       Because the bending doesn’t change along z axis,and the boundary conditions on top surface and under surface of sheet are: w ( ) 0, ( ) 0 2 2 = = = = t z z y t z x z  

薄城曲 0t=- o(少=Oy ax 将应力分量用挠度W表示的物理方程代入上式,并化 简得: aT Ez OVw az 1-u ax O 2y Ez a V-w az 由于挠度W不随z变化,且薄板在上下面的边界条 件为: (2)24=0,()=0 24

24 z y x z x y zy y xy zx x yx   −   = −     −   = −         将应力分量用挠度 表示的物理方程代入上式，并化 简得： w w y Ez z w x Ez z z y z x 2 2 2 2 1 1    − =      − =       由于挠度 不随z 变化，且薄板在上下面的边界条 件为： w ( ) 0, ( ) 0 2 2 = = = = t z z y t z x z  

Integrating the above two formulas on z, we get E 2t2)o 2(-2)4 V21 OX E ta ty=2(1-p 4)a From the third formula of differential equations of equilibrium, we get az:Ox ay Substituting into the expression which tEx, tyy are denoted by Bending w, and canceling terms, we get E a=2( -八(42)4 (1) 25

25 Integrating the above two formulas on z, we get: ( ) w x t z E zx 2 2 2 2 2 1 4            − − =   ( ) w y t z E zy 2 2 2 2 2 1 4            − − =   From the third formula of differential equations of equilibrium,we get: z x y z zx zy   −   = −     Substituting into the expression which are denoted by Bending ,and canceling terms, we get zx zy  , w ( ) z w E t z z 2 4 2 2 2 1 4         − − =     （1）