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

ElasticityChheuoeaem

1 Elasticity

The Solution ofthe Space ProblemChapter 8 The Solution of the Space Problem$8-1Solving the SpaceProblemAccording totheDisplacement$8-2The Semi-space Body Subjected to Gravity andUniformlyDistributedPressure88-3A Semi-space Body is Subject to NormalConcentration Force at the Boundaryg8-4 Solving the Space Problem According to the Stressg8-5The Torsion of Equal-Section Straight Pole88-6Membrane Analogy of theTorsion Problem88-7 The Torsion of Elliptic Section Pole$8-8 The Torsion of Rectangular Section Pole2

2 Chapter 8 The Solution of the Space Problem §8-5 The Torsion of Equal-Section Straight Pole §8-7 The Torsion of Elliptic Section Pole §8-6 Membrane Analogy of the Torsion Problem §8-8 The Torsion of Rectangular Section Pole §8-1 Solving the Space Problem According to the Displacement §8-2 The Semi-space Body Subjected to Gravity and Uniformly Distributed Pressure §8-3 A Semi-space Body is Subject to Normal Concentration Force at the Boundary §8-4 Solving the Space Problem According to the Stress

The Solutionof the Space ProblemMaterial mechanics has solved the torsion problems of roundsection pole, but it can't be used to analyze the torsion problemsof non-round section pole. For the torsion of arbitrary sectionpole, it is a relatively simple spatial problem. According to thecharacteristic of the problem, this chapter first gives thedifferential functions and boundary conditions, which the stressfunction should satisfy to solve the torsion problems. Then, inorder to solve the torsion problems of relatively complex sectionpole, we introduce the method of membrane analogy.3

3 Material mechanics has solved the torsion problems of round section pole, but it can’t be used to analyze the torsion problems of non-round section pole. For the torsion of arbitrary section pole, it is a relatively simple spatial problem. According to the characteristic of the problem, this chapter first gives the differential functions and boundary conditions, which the stress function should satisfy to solve the torsion problems. Then, in order to solve the torsion problems of relatively complex section pole, we introduce the method of membrane analogy

The Solution of the Space ProblemS 8-1 Solving the SpaceProblem Accordingtothe DisplacementThe strain component is represented by the deformationcomponent, and the physical equation can be rewritten as:EowEOuu72(1 + μ)aydx1+u1-2μEuOvEduH2(1 + μ)Ozax1 +a1u2AEavauEwuA2(1 +yaxOzu1+μ-2uEμOWhere 0=&.+&. +c.X1+ μ-2u1一4

4 §8-1 Solving the Space Problem According to the Displacement The strain component is represented by the deformation component, and the physical equation can be rewritten as: Where x y z     = + + ( ) 1 1 2 ( ) 1 1 2 ( ) 1 1 2 x y z E u x E v y E w z                 = + + −   = + + −   = + + −  ( ) 2(1 ) ( ) 2(1 ) ( ) 2(1 ) yz zx xy E w v y z E u w z x E v u x y         = + +     = + +     = + +   ( ) 1 1 2 x x E       = + + −

The Solution of the Space ProblemBy substituting the above equation into the equilibrium differentialequation, we get:E00atyx1agatruX=0f=0Ozaxay2(1+ μ) (1- 2μ axao,atyaty+f=0E100azayax+Vv=(2(1+μ) (1-2 μ yOtyao.atxf=0OzaxayE1a0VW2(1+ μ) (1- 2 μ oza2a2a?72+Qy2ax?az2The above equation is the basic differential equation for solvingspace problems by displacement.5

5 By substituting the above equation into the equilibrium differential equation, we get: ( ) 1 2 0 2 1 1 2 x E u f x       +  + =   + −    ( ) 1 2 0 2 1 1 2 y E v f y       +  + =   + −    ( ) 1 2 0 2 1 1 2 z E w f z       +  + =   + −    2 2 2 2 2 2 2 x y z  +   +    = The above equation is the basic differential equation for solving space problems by displacement. 0 0 0 x zx yx x y zy xy y z xz yz z f x y z f y z x f z x y              + + + =         + + + =          + + + =    

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: