东北大学材料成型力学考研课件(编辑修改稿)

东北大学材料成型力学考研课件(编辑修改稿)内容摘要：

   s              Neglect the terms in higher powers of x y zand    , , , , i j i j i ji j i jx x y y z z x y z x y zx y zs s ss    s             At the point A (O A). ,x y z  incremet :    , , , , i j i jBi j i j i jx x y y z x y z x yxysss s   s       At the point B (O B). ,xyincremet : x z y xzyA B C D E F G O At the point G (O G). xincremet :    , , , , ijGi j i j i jx x y z x y z xxss s  s     x z y xzyA B C D E F G O xsxyxzysyxyzzszxzyxx xxss xyxy xx xzxz xx yy yyss yxyx yy yzyz yy zz zzss zxzx zz  zyzy zz x z y xzyA B C D E F G O 0X Consider that the stress on each face are uniform (infinitesimal) 000XYZ   The element is in equilibrium xxxx x y zxss   zxzx z x yz   0yxx x z xx y zs    x z y xzyA B C D E F G O xsxyxzysyxyzzszxzyxx xxss xyxy xx xzxz xx yy yyss yxyx yy yzyz yy zz zzss zxzx zz  zyzy zz xx yzs   yxyx y x zy    yxxz  yx xy  00yxx x z xx y zs    0x y y y z yx y z s       0yzxz zzx y z s     0ijixs By means of tensor notation: j : is free subscript, it appears only one time in one term and is the same in all terms, it is replaced by x, y and z cyclicly in different eqs. i : is dummy subscript, it appears twice in one term and it is considered as the sum of three terms, in which the sub is replaced by x, y and z cyclicly . 0ijixs 0x j y j z jx y zx x xs s s      ,jx 0yxx x z xx y zsss    ,jy 0x y y y z yx y zs s s      ,jz 0yzxz zzx y zss s     ii x x y y zzs s s s  i ij jSlsdummy , ,。 j x y zi ix x iy y iz zS l l ls s s  ,ix x x x x x y y x z zS l l ls s s  y y x x y y y y z zS l l ls s s  ,iyz z x x z y y z z zS l l ls s s  ,iz Couple equilibrium (力矩平衡方程 ) Consider the moment about the axes passing through point P and parallel to Ox, Oy and Oz Neglect body forces and inertia forces. Resultant couple about the three axes: 000xyzMMM   0xM yzyz yyzyzy zzAfter neglecting the quantities of the fourth order and simplifying yz zyxz2y yz2yxy2z zy2z 0x z y xzyA B C D E F G O zyyxyx yy zyzy zz yzxzxySimilarly: 0yM  xz z x0zM  xy yxTherefore: ij jissEquality of shear stress（切应力互等） From above analysis we got conclusion that the stress tensor is a symmetric tensor. Therefore, nine ponents bee six ponents. 所以，应力张量是对称张量， 9个分量简化为 6个分量。 Three dimensional stress analysis (important feature of tensor) Resultant stress on an oblique plane inclined to three Cartesian axes Take an element using the method of sections Intersected by Three Cartesian plane Tetrahedron OABC (四面体 ) x z y O A B C xsxzxyys yzyxzszxzyRSnS sSzSxSySl m n RSzSySxS222R x y zS S S S  An oblique plane ABC （三维应力分析） （与三个坐标轴相倾斜的斜面上的和应力） Determine according to on three Cartesian planes: RS ijsplane area ox oy oz OBC l sxx xy xz OAC m yx syy yz OAB n zx zy szz ABC 1 Sx Sy Sz OABC in equilibrium 000XYZ   x x x y x zxS l m ns    From first column of the table y x y y y zyS l m n s   From second column of the table z x z y z zzS l m n  s   From third column of the table nmlllllzyxi iijj lS s j x j x y j y z j zx j y j z jS l l ll m ns s ss s s    from sub. j=x x x x y x zxS l m ns     j=y y x y y y zyS l m n s   j=z z x z y z zzS l m n  s  222R x y zS S S S  ijsilis known is known Sj is known SR Therefore, if the stresses on three orthogonal Cartesian planes are known, the stresses on any oblique plane can be determined. ijs can be transformed ji s Normal stresses on the oblique plane （斜面上的正应力） SR         r r r r oR SSSS       n x x x y x z y x y y y z z x z y z zS l m n l l m n m l m n ns    s    s         2 2 2 2x y z x y y z z xl m n lm m n n ls s s         Sn normal ponent, is coincide with ON Ss shear ponent resolve shear stresses on the oblique plane （斜面上的剪应力） 222 snR SSS  22s R nS S S Stresses boundary conditions （应力边界条件） Relations between the distribution load on the body surface and stresses within the body at the same boundary point are the stress boundary condition. （所谓应力边界条件是指物体表面上的应力分布与同一边界处物体内部应力之间的关系。 ） Body surface: inclined, normal ON ( direction cosines: l, m, n) Distribution load p [ px(ox), py (oy), pz(oz)] j ij iplsnmlp zxyxxxx s nmlp zyyyxyy s nmlp zzyzxzz s 0yxx x z xx y zs    0x y y y z yx y z s       0yzxz zzx y z s     0ijixs By means of tensor notation: （力平衡微分方程） ij jissEquality of shear stress（切应力互等） iijj lS sj x j x y j y z j zx j y j z jS l l ll m ns s ss s s    222R x y zS S S S  Summary of Last Class （斜面上的正应力）      n x x x y x z y x y y y z z x z y z zS l m n l l m n m l。