The Basic Theory of the Plane ProblemS 2-2Equilibrium Differential EquationPlane stress problem and plane strain problem are both the problems in planexy, and all the physical quantities have nothing to do with z.Below we discuss the relationships between stress of every points and bodyforce when the object is placed in the state ofequilibrium. And the equilibriumdifferential equations will be deduced. From the lamella shown in Fig.2-3,wetake out a small and positive parallelepiped PABC, and take for an unit length inthe directional dimension in z.Establishing the function of the positivestress force in an unit on the left sidex0iso, =o,(x,y) ,the coordinate on the right sideO2APx gets the increment dx , and the positiveaadxstress on the face is ,(x+ dx,y). SpreadingaxatOthe formula above by Taylor's series is:dxaxBatdo,(x,y)力yixdx+,(x+dx,y)=,(x,y)+ayaoyaxdyOay1 o(x,y)1 a",(x,y)(dx)"(dx)+.ax22!ax"n!Fig.2-36
6 §2-2 Equilibrium Differential Equation Plane stress problem and plane strain problem are both the problems in plane xy, and all the physical quantities have nothing to do with z. Below we discuss the relationships between stress of every points and body force when the object is placed in the state of equilibrium. And the equilibrium differential equations will be deduced. From the lamella shown in Fig.2-3,we take out a small and positive parallelepiped PABC, and take for an unit length in the directional dimension in z. y o x y dy y y y + x dx x x x + xy dx x xy xy + yx dy y yx yx + P A B C X Y D Fig.2－3 Establishing the function of the positive stress force in an unit on the left side is ,the coordinate on the right side x gets the increment , and the positive stress on the face is . Spreading the formula above by Taylor’s series is: (x, y) x = x dx(x dx, y) x + n n x n x x x x dx x x y n dx x x y dx x x y x dx y x y ( ) ( , ) ! 1 ( ) ( , ) 2! 1 ( , ) ( , ) ( , ) 2 2 2 + + + + = +
The Basic Theoryof the Plane ProblemAfter omitting small quantity of the two rank and above the two rank, we can)+o.( ax, and we get o, , fsy , w in a similar way. Then we get theget or(x,y)+axstress state shown as the Fig.2-3.While considering the body force in the plane stress state, the shear stressreciprocal theorem can be proved. Regarding straight line in parallel with theshaft of z and through the center D as the moment shaft, we can list theequilibrium equationof the moment ZM, =0:aTxydxdxdx)dyxT22axotdydy:0d)1C22ayBy dividing dxdy in the both sides of the above formula, we get:1 0ty1 Ot yxdydx =T.+xyyx2 ax2aySet dx → 0,dy →0, i.e. omitting small quantity, we have: T xy, = T,Lyx7
7 After omitting small quantity of the two rank and above the two rank, we can get , and we get , , in a similar way. Then we get the stress state shown as the Fig.2-3. dx x x y x y x x + ( , ) ( , ) y xy yx While considering the body force in the plane stress state, the shear stress reciprocal theorem can be proved. Regarding straight line in parallel with the shaft of z and through the center D as the moment shaft, we can list the equilibrium equation of the moment MD = 0 : 0 2 1 2 ( ) 1 2 1 2 ( ) 1 − = − + + + dy dx dy dy dx y dx dy dx dx dy x yx yx yx xy xy xy By dividing in the both sides of the above formula, we get: dxdy dy y dx x yx yx xy xy = + + 2 1 2 1 Set dx → 0,dy → 0 , i.e. omitting small quantity, we have: xy yx =
The Basic Theory of the Plane ProblemIn order to deducing the equilibrium differential equation ofthe plane stress problem, the equilibrium equation of the unit islisted as:ZF, =0:1at00 dx).dy×1-0, dy×1+(vx) . dx ×1a.+axay-t.·dx ×1 + X ·dx·dyxl = 0ZF, =0:dgatdxly)dx x1-o...dx ×1+yxayax-t:dyxl+Y.dx·dyxl=08
8 In order to deducing the equilibrium differential equation of the plane stress problem, the equilibrium equation of the unit is listed as: 1 1 0 ( ) 1 1 ( ) 1 0 : − + = − + + + = dx X dx dy dy dx y dx dy dy x F yx yx x yx x x x 1 1 0 ( ) 1 1 ( ) 1 0 : − + = − + + + = dy Y dx dy dx dy x dy dx dx y F xy xy y xy y y y
The Basic Theory of thePlane ProblemBy sorting them, we get:OTyxdgLX=0+axdydgOTxyy=0YayOxThese two differential equations include three unknownfunctions x,y, tx, = t x .Therefore, the problem of determiningthe stress components is indeterminate problem. In order tosolve this problem, the deformation and displacement must beconsidered.For the plane strain problem, the front and back faces stillhave , . But they do not affect the establishing of the aboveequations. So the above equations apply to two kinds of theplane problemsI9
9 By sorting them, we get: 0 0 + = + + = + Y y x X x y y xy x yx These two differential equations include three unknown functions .Therefore, the problem of determining the stress components is indeterminate problem. In order to solve this problem, the deformation and displacement must be considered. For the plane strain problem, the front and back faces still have . But they do not affect the establishing of the above equations. So the above equations apply to two kinds of the plane problems. z x y xy yx , , =
The Basic Theory of the Plane ProblemS2-3 The Stress State at a Point in a PlaneProblem1.The stress on the inclined planeHaving known the stress components x, y, Tx, = t yx of any point P insidethe elastic body, we try to get the stress which pass the point P on the arbitrarilyinclined cross section. From neighborhood of the point P we take a plane ABwhich is in parallel with the above inclined plane, and draws a small set squareor three column PAB on two planes which pass point P and is perpendicular tothe shaft of x and y. When the plane AB approaches point P infinitely, the stresson the plane AB will become the stress on the above inclined plane.0xSet the length of the surface AB in the plane0xy as dS, and N is the exterior normal direction.4PTand its direction cosine isxyacos(N,x)=l,cos(N,y) = mBAYNFig.2-410
10 §2-3 The Stress State at a Point in a Plane Problem 1.The stress on the inclined plane Having known the stress components of any point P inside the elastic body, we try to get the stress which pass the point P on the arbitrarily inclined cross section. From neighborhood of the point P we take a plane AB, which is in parallel with the above inclined plane, and draws a small set square or three column PAB on two planes which pass point P and is perpendicular to the shaft of x and y. When the plane AB approaches point P infinitely, the stress on the plane AB will become the stress on the above inclined plane. x y xy = yx , , Set the length of the surface AB in the plane xy as dS, and N is the exterior normal direction, and its direction cosine is: cos(N, x) = l,cos(N, y) = m P A B xy x y N yx N X N YN y S N x Fig.2－4 o