弹塑性力学第六章

6. ELASTIC PLANE PROBLEMS6.1. Basic equationsWe now proceed to a large category of problems of theory of elasticity, which are important for practical application and at the same time admit considerable simplification in the mathematical aspect of solution. Simplification implies that one may disregard one of the coordinate axes in these problems, for instance Oz , and consider that the whole phenomenon takes place in one plane Oxy. These problems fall into two groups opposite in a sense but common by the mathematical form of solution. The first of these groups is so called plane stress problems, and the second of these groups is called plane strain problems.The coordinates used in the following four sections are rectangular Cartesian, and we use Latin indices for range x, y, z as before and, in the plane problems, use Greek indices for range x , y .(a) Plane stress problemsIn this case, we have,0===ττσyz zx z (6.1) And),(y x x x σσ=, ),(y x y y σσ=, ),(y x xy xy ττ= (6.2) Hence, the equilibrium equations become0=+∂∂+∂∂F yx bx xy x τσ 0=+∂∂+∂∂F yx by y xy στ (6.3) The boundary conditions will bel l p xy x x 21τσ+=l l p y xy y 21στ+= (6.4)where p p y x , are the components of surface forces at x ,y directions and()()y n l x n l ,cos ,,cos 21== Constitutive equations become()y x x Eνσσε-=1 ()x y y Eνσσε-=1 ()τντγxy xy xy EG +==12 (6.5) and ()⎭⎬⎫+-===σσεγγy x z yz xz ,0 (6.6) where the strain component εz is in the direction z of the thin flat plate. The strain compatibility condition will bey x y x xy x y ∂∂∂=∂∂+∂∂γεε22222 (6.7) When the method of stress is used, the strain components in equation (6.7) must be exchanged for stress components. Considering equations (6.5) and (6.3) we have()()⎪⎪⎭⎫ ⎝⎛∂∂+∂∂+-=+⎪⎪⎭⎫ ⎝⎛∂∂+∂∂y F x F y x by bx y x νσσ12222 (6.8) When the body forces are constant ( or neglected ) throughout the volume of the body, then we have()02222=+⎪⎪⎭⎫ ⎝⎛∂∂+∂∂σσy x y x (6.9) or()02=+∇σσy x (6.9’) where ∇2is the Laplacian operator. The equation (6.9) is so-called M. Levy’s equation.Thus the solution of the plane stress problem (neglected the body forces) is reduced to integration of three differential equations:()y x ,,,0,==βασβαβ (6.3’) 02=∇σαα (6.9’)and satisfaction of the conditions on the surface:p n αβαβσ= (6.4’)(b) Plane strain problemsA body is said to be in the state of plane strain or plane deformation, parallel to th e xy -plane, if the component w of the displacement vector u vanishes and the components u and v are the functions of the coordinates x and y . Thus the plane strain problem is characterized by the formulas,()()y x w y x u u ,,0,=⎪⎭⎪⎬⎫==ααα (6.10)()u u αββααβε,,21+=(6.11) The constitutive relations are (),,,u u αββααβαβμδλθσ++= (6.12),0,===ττλθσyz xz z (6.13)where the dilatation .,u ααθ= it is easy to show that σz is proportional to the sum σσy x +:()()σσμλλσy x z ++=2 (6.14)The compatibility equation in stress form is()F b ααααμλμλσ,222++-=∇ (6.15) The boundary conditions arep n αβαβσ= (6.16)where the n βare components of the exterior unit normal vector to boundary curve C. The formulation of the first boundary-value problem is complete.It is interesting to note that in the case of constant body forces the equations determining stress distribution do not contain the elastic constants of the material. Hence the stress distribution is the same for all isotropic materials, provided the equations are sufficient for the complete determination of the stress. This conclusion is of practical importance. It is possible to determine stress by an optical method using polarized light. From the above discussion it evident that experimental results obtained with a transparent material can be applied immediately to any other material, such as steel and other metals. It should be noted also that in the case of constant body forces the compatibility equation holds both for the case of plane stress and for the plane strain. Hence the stress distribution is the same in the two cases provided the shape of the boundary and external forces are the same.6.2. Stress function. Inverse and semi-inverse methodIt has been shown that a solution of two-dimensional problems reduces to the integration of the differential equations of equilibrium together with the compatibility equation and the boundary conditions. If we neglected the body forces, the equations to be satisfied are⎪⎪⎭⎪⎪⎬⎫=∂∂+∂∂=∂∂+∂∂00y x y x y xy xy x σττσ (6.17) ()02222=+⎪⎪⎭⎫ ⎝⎛∂∂+∂∂σσy x y x (6.18) and the boundary conditions (6.4)For the first boundary-value problems, i.e. given the surface forces, the usual method of solving these equations is by introducing a new function, called the stress function introduced by G . B. Airy (1862), and is sometimes called the Airy stress function . Let us take any function ψ:()y x ,ψψ=and put the following expressions for stress components:⎪⎪⎪⎪⎭⎪⎪⎪⎪⎬⎫∂∂∂-=∂∂=∂∂=y x x y xy y x ψτψσψσ22222 (6.19) Substituting expressions (6.19) into Eq. (6.17) it is complete satisfied. The true solution of the problem is that which satisfies also the compatibility equation (6.18). Substituting expressions (6.19) into (6.18) we find that the stress function ψ must satisfy the equation024422444=∂∂+∂∂∂+∂∂y y x x ψψψ (6.20) or04=∇ψ (6.20’)Equation (6.20) is called a biharmonic equation. Thus the solution of a two-dimensional problem, when the body forces are neglected, reduces to finding a solution of equation (6.20), which satisfies the boundary conditions (6.4) of the problem.Now we consider a more general case of body forces and assume that these forces have a potential. Then the components F b x and F by are given by the equationsyV F x V F by bx ∂∂-=∂∂-=, (6.21) where V is the potential function. Equation of equilibrium become()0=∂∂+-∂∂xV x xy x τσ ()0=∂∂+-∂∂xV y xy y τσ These equations are of the same form as equations of equilibrium and can be satisfied by taking⎪⎪⎪⎭⎪⎪⎪⎬⎫∂∂∂-=∂∂=-∂∂=-y x x V y V xy y x ψτψσψσ22222, (6.22) Substituting (6.22) in the compatibility equation (6.8) and (6.15) respectively, we have: for plane stress problem,()V ∇--=∇241νψ (6.23) for plane strain problem,V ∇---=∇24121ννψ (6.24)In many problems it is often found convenient to employ the inverse method or semi-inverse method. In the inverse method, some functions satisfying the differential equations are taken and examined to see what boundary conditions these functions will satisfy and thereby to know what problems they can solve. Specifying in advance the analytic form of stress function ()y x ,ψ and selecting its parameters (for instance, coefficients of a polynomial) so as to satisfy the basic equations and boundary conditions for plane problems. The semi-inverse method in which one guesses at part of the solution, tries to determine the rest rationally so that all the differential equations and boundary conditions are satisfied. As we known, from uniqueness of solution, any by guessing solution, which can be satisfied all the differential equations and boundary conditions of a well-posed plane problem is an exact solution of this problem.As we know, the biharmonic equation is of fourth-order differential equations, therefore all of polynomials whose order is lower than fourth are biharmonic function. When the stress function is taken as a polynomial in linear form,()cy bx a y x ++=,ψ, (6.25)the compatibility equation is satisfied for all values of the constant a, b and c. The stress components given by Eqs. (6.19) are zeros. The stress boundary conditions always give zeros. Thus, we have the case of no surface forces and no stress and that the superposition of a linear function to the stress function for any problem does not affect the stresses. If the function ψ represents a polynomial of the second degree()y c bxy x a y x 2222,++=ψ (6.26) equation (6.20) will obviously be satisfied everywhere for any values of a, b, and c. The stresses, in accordance with equations (6.19) will be expressed as.;;b a c xy y x -===τσσ (6.27)Then we have the case of the homogeneous state of stress.If the stress function ψ is a polynomial of third degree()y c bxy x a y k y x f y x e x d y x 223223226226,++++++=ψ (6.28) equation (6.20) will again be satisfied for arbitrary values of the coefficients the stresses, according to equations (6.19), will represented as⎪⎭⎪⎬⎫---=++=++=b fy ex a cy dx c ky fx xy y x τσσ (6.29)i.e., they will be linear functions of the coordinates.If the stress function ψ is given as a function of the fourth or higher degree, its derivatives entering in equation (6.20) will be different from zero, in general; the coefficients must be chosen in such a way as to satisfy the compatibility condition (6.20) for arbitrary values of x and y . The stresses will then be functions of the second and higher degree.6.3. Bending of Elastic BeamsAs a example, we now discuss the bending of elastic cantilever beam loaded at the free end, as shown in figure (6.1), by using the semi-inverse method and stress function. Consider a cantilever having a narrow rectangular cross section of unit width bent by a force F applied at the free end δis the thickness of the beam in the direction perpendicular to the plane of the drawing. The upper and lower edges are free from load, and the shearing forces, having a resultant F , distributed along the end x=0. In this case , the boundary condition are⎪⎪⎭⎪⎪⎬⎫-====⎰+-±=±==h h xy h y y h y xy x x dy F δτστσ0)(0)(0)(0 (a)(a) Selecting stress functionLet us try to solve this problem by using the semi-inverse method. As we known from the mechanics of materials, The moment will be changed linearity along x and the normal stress σx is proportional to y. Thus, it can be supposed thatxy C y x 122=∂∂=ψσ (b) where C 1 is a constant. Integrating the equation (b) twice, we have()()()x f x f y y x C y x 213161,++=ψ (c)Differentiating this function with respect to x and substituting the results in the biharmonic equation (6.20), we have0424414=+dx f d dx f d y (d) The second term in the above equation is independent of y, but this equation must be satisfied for all values of x and y in the beam. This is possible only if0414=dxf d and 0424=dx f d (e) or C x C x C x C f 5423321+++=C x C x C x C f 9827362+++=where C C C 932,,,⋅⋅⋅ are constants of integration. Therefore, we have()()C x C x C x C C x C x C x C y y x C 9827365423323161++++++++=ψ (6.30)Substituting in Eqs. (6.19), we obtain Fig., 6.1 A cantilever beam loaded at its free end()()C y C x C xy C x y 73622226+++=∂∂=ψσ (f) C x C x C y C y x xy 43222122321----=∂∂∂-=ψτ (g) Determining the coefficientsThe boundary conditions require that ,0h y on y ±==σ or()()0267362=+++C h C x C h C()()0267362=+-++-C h C x C h CThese equations must be valid for all values of x between 0 and l ; it follows therefore that062=+C h C , 073=+C h C0,07362=+-=+-C h C C h CSolving, we find .07632====C C C C Thus()02321432221=----=±=C x C x C h C h y xy τ orh C C 21421-=. From the fourth equation of (a), we have()F dy h y dy h h h h xy =-=-⎰⎰--22δδτ, from which,2331I F h F C -=-=δ where h I 332δ= is the moment of inertia of the cross section, the final expressions for the stress components are therefore()⎪⎪⎭⎪⎪⎬⎫--==-=y h I F I F x y xy y x 2220τσσ (6.31)This coincides completely with the elementary solution as given in the mechanics of materials. From this solution, we see that the distribution of the shearing stress at the ends must according to the parabolic law and that σx at the built-in end must vary linearly with y . The solution, therefore, is exact only when the boundary forces are so given. If the boundary forces are given in any other manner, this solution will not be an exact one; but by virtue of Saint-Venant’s principle, it does represent the stress distribution for some cross section at a large distance from the ends.Displacements calculationNow we discuss the deformation of the beam. From the strain-displacement relations and Hooke’s law, we obtain,,EIFxy y v EI Fxy x u ν=∂∂-=∂∂ ()()EIy h F E x v y u xy 22)1(12-+-=+=∂∂+∂∂νντ(h)Integrating the above first two equations, we obtain ()()⎪⎪⎭⎪⎪⎬⎫+=+-=x EI y Fx v y u EI y x F u i 12222νν (i) in which ()()x v y u 11, are the unknown functions of y and x. substituting u and v into Eq. (h) we find()()h F EIx EI F dx dv y EI F dy du 221211222νν+-+-=+- We note that the terms on the left of the equal sign are functions of y alone and the terms on the right side are functions of x alone. A function of x can be equal to a function of y for all values of x and y only when they are both equal to a constant sayC 1. Thus()C y EIF dy du 12122=+-ν()C h F EIx EI F dx dv 122112-=++-νIntegrating, we have()()C y C y EIFy u 213126+++=ν ()()C x C h Fx EIEI x F x v 3123116+-+-=ν where C 2 and C 3 are constants of integration. The displacements u and v are therefore()C y C y EI Fy x EI F u 2132262++++-=ν ()C x C h x EIFx EI F y x EI F v 31232162+-+-+=νν Assume that the right end (at the point x=l, y=0 ) of the beam is fixed. The boundaryconditions are then.0,0===∂∂==y l x at yu v uWe find by substitution,()h l EIF EI l F C C EIl F C 23322113,0,2ν++=== then the displacements are()()()()()⎪⎪⎭⎪⎪⎬⎫⎥⎦⎤⎢⎣⎡-++-++=++-=x l h l y x l x EI F v EIy F y x l EI F u ννν123662222233322 (6.32) The functions of u and v are non-linear equations of x and y . It means, that any sectionwill be not a plane after deformation. This phenomenon is different with the result of elementary theory of mechanics of materials. At the fixed end (x=l ), the deformation will be given by the following relations:()⎪⎪⎭⎫ ⎝⎛-+==222323l ly l EI F v lx ν ()⎥⎦⎤⎢⎣⎡+-=⎪⎭⎫ ⎝⎛∂∂=νν122h y EI F x v l x ()GI h F EI h F x v y l x 21220-=+-=⎪⎭⎫⎝⎛∂∂==ν Then we obtain the angle of rotation of the horizontal line element at the fixed end isFh 2/2GI ( Fig. 6.2). Let us now realize the same fixing in a different manner specifying the condition0,0,=∂∂==xvandy l x which requires that the element dx of the supporting section must remain horizontal, then we obtain the same result as the preceding discussion, i.e. the angle of rotation of the vertical line element at the fixed end is also Fh 2/2GI.The vertical displacement of the axes of the beam is()()()x h EIF EI Fl EI x l F EI x F v y -+++-==1132623220ν (6.33) The deflection at the free end is()()GIh Fl EI FlEI l Fh EI Fl v y x 231323230+=++===ν (6.34) The second term of the above equation is obviously the influence to the shearing stress,but it is too small , about ( 2h/l )2. For example, if l =10(2h), then the ratio( Fh 2/2DI over Fl 3/3EI) is 1/100.Fig. 6.2 Theoretical results of the fixed endIt is clear, the result of mechanics of material is sufficient precisely for the long thin beams.In fact, detail analysis of the deformation at the end is very complex, but the influence region of end effect, according to the Saint-venant principle, is limited (about like as the height of the beam).Example 6.1 Find the stress distribution and the deflection at center of the beam subjected uniformly distributed load q along its top edge as shown in Fig. 6.3.(1) Select the stress function⎪⎪⎭⎫⎝⎛-+++=55324332221y y x C y C y x C x C ψ (6.35)Easy to verify that the stress function ψ satisfy the biharmonic function.(2) Determine the coefficientsThe boundary conditions are()0=±=h y xy τ ()q h y y -=-=σ ()0==h y y σand at x=,l ± we have 0=⎰-ydy hh x σδτql dy l hhxy ±=⎰-Then we findC 2 + 3C 4h 2 =02(C 1 + C 2h +C 4h 3)=0Fig. 6.3 Simple supported beam subjected uniformly distributed load.2(C 1 – C 2h – C 4h 3 )=-q C 3 +C 4l 2 – (2/3)C 4h 2 = 0 4(C 2h + C 4h 3 ) =qSolving the above equations we obtain⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛-++-=5815288345323223322y y x h h l h y h y x x q ψ (3) Find the stress components()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=53222222h y y x l I q x σ (6.36) ⎪⎪⎭⎫⎝⎛+--=h y h y Iqy 3233232σ (6.37) ()x y h Iq xy 222--=τ (6.38) where h I 332δ=. Comparing with the results of mechanics of materials, we find the second term of equation (6.36) is independent on x , and when h = l/10, the influence upon σx is very small (about 1/1500). It should be noted that at ,l x ±=()153522h y I qy x -=σ and not coincide with the given conditions, i. e. at eachend, the normal stresses are not equal to zero. But the resultant force and resultant moment are equal to zero. Again according to the Saint-Venant’s principle the results of stress analysis are correctly for the whole beam.(4) determine the deflection at the center section of the beamBy using the above results we can find the deflection v 0 at y=0:⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛++--=x h x h x x l EI q v 222242202151222ν (6.39) and the curvature of the deflection curve is⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛++-=h x l EI q dx v d 2222022542ν (6.40)It is easy to understand that the second term of Eq. (6.40) is a correction term..Fig. 6.4 A part loaded beam6.4. Deep beam problem Solution by trigonometric seriesThe method of solution of the plane problem by means of algebraic polynomials, considered in the preceding sections, afford limited possibilities for practical application because it is very difficult to select a polynomial which gives the solution corresponding to a prescribed load which is more or less complicated. The method of trigonometric polynomials, proposed by C.Ribie`re (1889) and L.N.G. Filon (1903) for the case of bending a rectangular strip whose length is considerably greater than the height, was found to be far more effective. Specially, when the external load is discontinuity, select a trigonometric polynomial as a stress function is reasonable.We now discuss the solution by trigonometric polynomial, as a example of part loaded beam (Fig.6.4), by using the inverse method. Select the stress function as the following form()()lxn y f y x πψsin,= (6.41) where n is an arbitrary constant, f(y) is an arbitrary function, 2l is the length of the beam.As indicated in the preceding sections, the solution of the plane problem by means of the stress function ()y x ,ψ is reduced to the integration of equation04=∇ψLet ln πα=, then we have ()x y f αψsin =()ααψsin 444y f x=∂∂ ()xy f yx ααψsin 2224''-=∂∂∂()()x y f yαψs i n 444=∂∂ then we obtain a ordinary differential equation of f(y)()()()()02424=+''-y f y f y f ααIts general solution is in the usual way in the hyperbolic function form()y ysh C y ych C y sh C y ch C y f αααα4321+++= (6.42)Substituting (6.42) in (6.41), we find the stress function as()x y ysh C y ych C y sh C y ch C αααααψsin 4321+++= (6.43)The stress components are()()]22[sin 43222122y ysh y ch C y ych y sh C y sh C y ch C x y x αααααααααααααψσ+++++=∂∂=()y ysh C y ych C y sh C y ch C x xy ααααααψσ4321222sin +++-=∂∂= (6.44)()()][cos 43212y ych y sh C y ysh y ch C y ch C y sh C x yx xy ααααααααααααψτ++++++-=∂∂∂-=(6.45)It is easy to verify that another form of hyperbolic function will be satisfied the biharmonic function of ψ. Therefore we can get infinite particular solution for this problem, hence the stress function may be taken as the following form ()()∑∞=+++=14321s i n,n y y c h C y y s h C y ch C y sh C x y x αααααψ+()∑∞=++++18765cos n y ych C y ysh C y ch C y sh Cx ααααα (6.46)The coefficients of equation (6.46) C i (I=1,2,…,,,8) will be determined by using theboundary conditions. In this case, the stress boundary conditions must be given in the form of infinite series:()x B x A A x q n nn nααsin cos 110∑∑∞=∞=++= (6.47)As we know from the mathematical analysis, any function in an interval []l l , can be expanded by Fourier series, and the coefficients of this series will be()()()⎪⎪⎪⎭⎪⎪⎪⎬⎫===⎰⎰⎰---l l n l l n l ldx l x n x q l B dx l x n x q l A dx x q l A ππsin 1cos 1210 (6.48) For the case of an uniformly loaded beam (q (x) = q 1), as shown in Fig. (6.4), theFourier coefficients are ()laq dx l q dx x q lA a a ll110221===⎰⎰--()a lq xdx x q l A ll n αααsin 2cos 11==⎰- ()⎰⎰--===a a ll n xdx lq xdx x q l B 0sin sin 11αα from which in the upper side of the beam, we have()⎪⎭⎫⎝⎛+=∑∞=11cos sin 22n a l a l a q x q ααα (6.49)For the lower side, we have the same representative. Now we, as a example, discussthe deep beams.Solution of continuous deep beamConsider a continuous deep beam consisting of a sequence of equal spans similarly loaded such a uniform pressure q on the upper edge. Let h and 2l be the height and span length respectively. In this case, the elementary beam theory is not adequate, but engineering useful results can be obtained by the method of trigonometric series.Let us neglect the influence of left and right neighborhood spans to the center span which we considered only (see Fig. 6.5).The supports of the continuous deep beam are series columns, now we simplified them as a series concentrated forces 2ql. Now we try to solve it. (1) Select the stress functionConsidering the sine function is an anti-symmetric function and σσy x , must be symmetric for y-axes, we select the stress function as the following form()y D xy D x D y ych C y ysh C y ch C y sh C x n 2322114321cos ++++++=∑∞=αααααψ (6.50)where D = const..l n πα=Obviously, due to ()x l x ααcos 2cos =+, the ‘center’ span is truly a substitute span.(2) Calculate the stress components()()D y sh C y ch C y ych C y ysh C y ch C y sh C x y n x 3434321122]2[cos ++++++=∂∂=∑∞=αααααααααψσ (6.51,a)()1432112222cos D y ych C y ysh C y ch C y sh C x x n y ++++-=∂∂=∑∞=ααααααψσ (6.51,b)Fig. 6.5The substitute span of a continuous deep beam=∂∂∂-=yx xy ψτ2()[]D y ych C y sh C y ysh C y ych C y sh Cy ch C x n 24343211sin -+++++∑∞=ααααααααα(6.51,c)The shearing stress components must be anti-symmetric, therefore we have D 2 = 0(3), Give the equilibrium conditions and boundary conditions a), ,0=τxy when, x=0 and x=l ; b),;0ql dx ly-=⎰σc), ;2,0,0,0,0l x x and y when y xy ≠≠===στ d), ;,,0h y when q y xy =-==στe),,00=⎰dy hxσ for any vertical section;(4), Determine the constants and find the stress distribution ruleEasy to find the constants D 1, D 2, D 3, C 1,……C 4,D 1= -q/2, D 2= 0, D 3= 0C 1= ααααααqhh hhch sh h q s 222-≈+-C 2 =α22qC 3= αααααqhh h hh qs s 222222-≈--C 4=ααααααqhh h hhch sh h q s 22222≈-+Substituting above expressions into (6.51) and considering the following related expressioneyy sh y ch ααα-=-we have()()()⎪⎪⎪⎭⎪⎪⎪⎬⎫-=-+-=--=∑∑∑∞=-∞=-∞=-111sin 21cos 21cos 2n y xy n y y n y x e y x q q e y x q e y x q αααααταασαασ(6.52)The stress distribution rule is shown as figure (6.6)(5), Determine the displacement componentsSubstituting the stress components into the following equations()σνσεy x x Ex u -==∂∂1()σνσεx y y Ey v -==∂∂1and integrating, we can obtain the displacement components u and v6.5 Basic equations in polar coordinatesUp to now, in solving the problems of the theory of elasticity we have used therectangular coordinates. In many problems, however, it is found more convenient to select other systems of coordinates. In particular, consider polar coordinate in planeproblems. For example, in discussing stresses in circular rings and disks, curved beams,Fig. 6.6 Stress distribution rule in deep beametc.The relations between polar and rectangular coordinates are (Fig. 6.7),sin ,cos ϕρϕρ==y x (6.53)xyy x arctan,222=+=ϕρ (6. 54) (1) Equilibrium EquationsLet us now consider the equilibrium of a infinitesimal element abcd cut out from the plate by the radial section od, oc , normal to the plate, and by two cylindrical surface ab, cd , normal to the plate. The normal stress component in the radial direction is,σρ the normal component in the circumferential direction is σϕ, and theshearing stress component is τρϕ, each symbol representing stress at the point ϕρ,. The body forces in the directions ϕρ,are F F b b ϕρ,, then from the equilibrium condition in the radial direction we have (Fig. 6.8)()-⎪⎪⎭⎫ ⎝⎛∂∂+--+⎪⎪⎭⎫ ⎝⎛∂∂+2sin ϕρϕϕσσϕρσϕρρρρσσϕϕρρρd d d d d d d 02cos 2cos 2sin=+-⎪⎪⎭⎫ ⎝⎛∂∂++ϕρρϕρτϕρϕϕττϕρσρρϕρϕρϕϕd d F d d d d d d d bFig. 6.7 Relations between polar and Fig.6.8 Notationsrectangular coordinateConsidering ϕd is so small, we have12cos,22sin≈≈ϕϕϕd d d Neglecting small quantities of higher order, and dividing through by the elementary area ,ϕρρd d we obtain the equation of equilibrium in the radial direction, and in a similar manner, the equation of equilibrium in the tangential direction can be obtained as following. Then the two equilibrium equations for plane problems in polar coordinates are⎪⎪⎭⎪⎪⎬⎫=++∂∂+∂∂=+-+∂+∂∂02101F F b b ϕρϕρϕϕρϕρρϕρρτρτϕσρρσσϕτρρσ (6.55)If the body forces are neglected, we observe that Eq. (6.55) will be identically satisfiedwith the introduction of the stress function ψ1 defined byϕψρρψρσρ∂∂+∂∂=22211 (6.56,a) ρψσϕ∂∂=22(6.56,b) ⎪⎪⎭⎫⎝⎛∂∂∂∂-=∂∂∂-∂∂=ϕψρρϕρψρϕψρτρϕ11122 (6.56,c) (2) Strain-displacement relationLet us consider the deformation of an infinitesimal element ABCD, and denote the radial and circumferential components of displacements in directions ϕρ, by u and v respectively (see Fig. 6.9).In the figure (6.9), the solid element ABCD is the state of before deformation, and the dot-line element A’B’C’D’ is the state of after deformation. The vector of displacement at a point can be divided into two parts: radial and circumferential components. For example, vector AA’ will be divided into vectors AA’’ and A” A’, etc. Here AB =ρd , the extension line of OA ’ intersect B’’B’ at F, A’E parallel to A’’B’’.1In this chapter we use ψas the stress function. Don’t confusion with the coordinate .ϕThe circular curve whose radius is OA ’ intersect OD’’ at G . The angle ∠ HA’F =π/2. Thus, we have the displacements at points A, B, C, D, asAA’= AA’’ + A’’A’; BB’= BB’’ +B’’B’;………..etc.Considering ϕd is a infinitesimal quantity, we can obtained the strain components:ρρρρερ∂∂=-∂∂+=''-''≈-''=u d ud uu ABA AB B AB AB B A=-'''-''+'''≈-''=ADADA A A G D D AD AD D A εϕ=()ρϕρϕρϕρϕρϕϕu d v d d v d u d d dvv +∂=--+++=''+''''-'≈''∠+''∠=+=H A HD B A EFE B H A DF A B γγγρϕ21=='''-''+'''''-'''''HA A A D D A O A AB A E B ==∂∂+-∂∂+++-∂∂+∂∂ϕϕϕρϕϕρρρρρρd v d ud u u u v d u d d v=.ϕρρρ∂∂+-∂∂u v v Fig. 6.9 Deformation of an infinitesimal fan element。