《弹性力学》课程教学课件（英文讲稿）Chapter 7 The Basic Theory of the Plane Problem

ElasticityChapter 7The Basic Theory of the Plane Problem

1 Elasticity

The Basic Theory of the Space ProblemChapter7TheBasicTheory of the SpaceProblemS 7-1 Equilibrium DifferentialEquationS 7-2 The Stress State at a Point in a Plane ProblemS 7-3 The Principle Stress.The Maximum and the Minimum StressS 7-4 Geometrical Equation. Physical EquationS 7-5 The Basic Equation of the Axisymmetric Problem2

2 Chapter 7 The Basic Theory of the Space Problem §7-4 Geometrical Equation. Physical Equation §7-3 The Principle Stress. The Maximum and the Minimum Stress §7-2 The Stress State at a Point in a Plane Problem §7-1 Equilibrium Differential Equation §7-5 The Basic Equation of the Axisymmetric Problem

The Basic Theory of the Space ProblemS 7-1 Equilibrium Differential Equation20idzazatsyTay+daAt any point P in theaatdzobject, take a smallparallelepiped shownatyeOtxas the figure. TheTya+-dytx+dxayaxatrVExa0stress components onTry+dxaxatyadyeach surfaceof aaortyx+TyaayOdxaxsmall parallelepipedshown in thearefigure..图7-13

3 §7-1 Equilibrium Differential Equation At any point P in the object, take a small parallelepiped shown as the figure. The stress components on each surface of a small parallelepiped are shown in the figure

The Basic Theory of the Space ProblemIf the line connecting the front and back00zof the hexahedron is AB, from mAs = OdzO.OzOydzwe can get:Tzy +O2at二atydydzattaxdydyOzyzdxdzoydydxdzBT+TyzL22ayaoydOyoOtaydzdzdz0dxdy+TTyAzy22OzdyTux+2Simplifying and removing the higher order smallamount, we have: Ty, = Tay

4 If the line connecting the front and back of the hexahedron is AB, from we can get: 0 mAB = d d d d d d d 2 2 yz yz yz y y y x z x z y         + +    d d d d d 0 2 2 zy zy zy z z z x y dxdy z       − + − =      Simplifying and removing the higher order small amount, we have: yz zy   = A B

The Basic Theory of the Space Problemaadz2ats-1a-atd-atdrTie+odyTrdxaxatnYtxtdadCadyaxarda.diSelayCd2T.Tx= Tx=TIn the same way:xzxvyxThe shear stress reciprocal relationship in general space are provedagain.5

5 In the same way: The shear stress reciprocal relationship in general space are proved again. zx xz xy yx = =    

The Basic Theory of the Space Problem00zdzO2OzatzydzLOzaadztzxO2aoydyOyayatyxdytyxoyZF,=0aoataydz)dy× dxdy).dx×dz -o,dx ×dz +(tayOzatxydx).dyxdz -txy dz×dy+ f, .dxdyxdz = 0-t.y·dxxdy+(TXax6

6 ( d ) d d d d ( d ) d d d d ( d ) d d d d d d d 0 y zy y y zy xy zy xy xy y y x z x z z y x y z x y x y z z y f x y z x            +   −   + +      −   + +   −   +    =  0 F y =

The Basic Theory of the Space ProblemZF =0,ZF, =0,ZF =0FromThree equations are listed and simplified:atagOTzxVXf =0OxOzdagotaTXyf.=0axOzOyot.tVz=0f-QzaxayThis is the equilibrium differential equation in rectangularcoordinates of space

7 From    F F F x y z = = = 0, 0, 0 Three equations are listed and simplified: This is the equilibrium differential equation in rectangular coordinates of space. 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 Basic Theory of the Space ProblemS 7-2 The Stress State at a Point in a Plane Problemzi1. Stress on an oblique sectionGPzZ.1OzCNKTNyaTXTTyz9yTxPyNsLg,Txy3福VBTixOyxZPx→a.BAyOzxA[=cos(N,x)n=cos(N,z)m=cos(N,y)xSAOAB=nSSsOAc=mSSAABC=SSAOBC-IS8

8 A B C x y z O 1. Stress on an oblique section x y z x  xy  yx z y  xz zx zy  yz O A B C y  yx  yz z zy zx  xy  xz x px py pz N l=cos(N,x) m=cos(N,y) n=cos(N,z) SABC=S SOBC=lS SOAC=mS SOAB=nS §7-2 The Stress State at a Point in a Plane Problem

The Basic Theory of the Space ProblemFx=0Px·S-ox·IS-tw·mS-tx·nS=0z1Px =lox +mtyx +ntzxGPzP, = ltxy + mo, + ntzyNP, =ltxz + mtyz +nozKyx0gySet the normal stress on thePyTxzinclined plane to be on, then3zBKTxZIPxO= pl+p,m+p,n→Aa.Substitute px, Py, p, into the abovexSAABC=SSAOBC=ISformula, we have:SsOAc=mSSAOAB=nS9

9 SABC=S SOBC=lS SOAC=mS SOAB=nS Fx = 0 px  S − x  lS − yx  mS − zx  nS = 0 x x m yx n zx p = l +  +  y xy m y n zy p = l +  +  z xz m yz n z p = l +  +  A B C x y z O y  yx  yz z zy zx  xy  xz x px py pz N Set the normal stress on the inclined plane to be σN, then N x y z  = + + p l p m p n Substitute px , py , pz into the above formula, we have:

The Basic Theory of the Space ProblemO~ =o,? +0,m2+0,n2+2mnt +2lnt x +2lmt.tn?=p?+p,+p?-o?When the oblique plane is the boundary, we can obtain the stressboundary condition as follows:lox +mtyx +ntx = JVx=1lt..+mo,+nt xyyV=lt.+mt. + no.xz J、J,、 J. are the surface force components on the boundarycondition.10

10 2 2 2 2 2 2 N x y z yz xz yx        = + + + + + l m n mn ln lm 2 2 2 2 2 N x y z N   = + + − ppp x yx zx x l m n f    + + = xy y zy y l m n f    + + = xz yz z z l m n f    + + = ❖ When the oblique plane is the boundary, we can obtain the stress boundary condition as follows: ❖ 、 、 are the surface force components on the boundary condition. x f y f z f