《弹性力学》课程教学课件（英文讲稿）Chapter 8 The Solution of the Space Problem

The Solution of the Space ProblemFor the space axisymmetric problems:up=u(p,z),u,= O,u, = u(p,z)o= To=0=00=0T2OopThe eguilibrium differential equations:at.ag0C0ZOC（XOzappatdgTpzOZ=0opOzp6

6 For the space axisymmetric problems: 0 z f z              − + + + =   0 z z z z f z          + + + =   The equilibrium differential equations: ( , ), 0, ( , ) z u u z u u u z   = = =   0  z   = = 0  0  = z  =

The Solution of the Space ProblemGeometric equations:Physical equationsOuEauuPap1+μap1-2μwpEuuCSHQ01+ μ1-2μppEOuuOu.HOC1+μOz1-2μOzOuEouououT0zp2(1 + μ)OzapzpOzap

7 Geometric equations: u     =  u    = z z u z   =  z z u u z       = +   1 1 2 E u           = +   + −    2(1 ) z z E u u z          = +   +     1 1 2 E u          = +   + −   1 1 2 z z E u z         = +   + −    Physical equations:

The Solution ofthe Space ProblemSo we can get the basic differential equation of space problemaccording to displacement:E0012(1 + μ)1-2μopE102(1 + u)1-2uOzouuOupL一apQzp2aa1 azappop8

8 So we can get the basic differential equation of space problem according to displacement: 2 2 2 1 0 2(1 ) 1 2 1 0 2(1 ) 1 2 Z z E u u f E u f z                 +  − + = + −         +  + = + −    z u u u z        = + +   2 2 2 2 2 1    z     = + +   

The Solution ofthe Space ProblemS 8-2The Semi-space Body Subjected toGravity and Uniformly Distributed PressureThe mass per unit volume of a semi-space body is p, which issubjected to uniformly distributed pressure on the horizontalboundary. The displacement boundary condition is w z=h = 0Ask for the displacement component and the stress componentSolution: Body force:fx = f,= 0, f, = pgSurface force:: z=0:=F,=0,J=qu= v= 0,w = w(z)Displacement:OvQuOwdwA++0dzOzaxayxd?wdz29Z

9 §8-2 The Semi-space Body Subjected to Gravity and Uniformly Distributed Pressure u = v = 0,w = w(z) Solution: q x z z=h o f x = f y = 0, f z = g 0: 0, x y z z f f f q = = = = Body force: Surface force: Displacement: z w y v x u   +   +    = d d w z = 2 2 2 d d w w z  = The mass per unit volume of a semi-space body is ρ, which is subjected to uniformly distributed pressure q on the horizontal boundary. The displacement boundary condition is . Ask for the displacement component and the stress component. w z=h = 0

The Solution of the Space ProblemSubstituting into the equilibrium differential equationqGao+GV?u+fx=01-2μ axOxa0G+GVv+ f, = 0z=h1-2μayGa0+GV2w+f,=01-2μOz7QuavOwA+OzaxayThe first two equations are naturally satisfied.dwand the third equation isdz2(1- μ)G dwd?w-pgdz21-2μdz210

10 q x z z=h o 0 1 2 2 +  + =   − x G u f x G   0 1 2 2 +  + =   − y G v f y G   0 1 2 2 +  + =   − z G w f z G   z w y v x u   +   +    = d d w z = 2 2 2 d d w w z  = 2 2 2(1 ) d 1 2 d G w g z    − = − − The first two equations are naturally satisfied, and the third equation is: Substituting into the equilibrium differential equation: