弹塑性力学复习思考题

1、研究生弹塑性力学复习思考题1.简答题：（1）什么是主平面、主应力、应力主方向？简述求一点主应力的步骤？（2）什么是八面体及八面体上的剪应力和正应力有何其特点（3）弹性本构关系和塑性本构关系的各自主要特点是什么？（4）偏应力第二不变量 J2的物理意义是什么？（5）什么是屈服面、屈服函数？ Tresca屈服条件和Mises屈服条件的几何 与物理意义是什么？（6）什么是Drucker公设？该公设有何作用？（能得出什么推论？）（7）什么是增量理论？什么是全量理论？（8）什么是单一曲线假定？（9）什么是平面应力问题？什么是平面应变问题？在弹性范围内这两类问题之间有和联系和区别？（10）论述薄板小挠度弯曲

2、理论的基本假定？二、计算题1、For the following state of stress, determine the principal stresses and directions andfind the traction vector on a plane with unit normal n = (0,1,1)2。311曾=10212 02、In suitable units, the stress at a particular point in a solid is found to be21-4巧j=140H01Determi ne the tracti on vect

3、or on a surface with un it no rmal (cos ds in ,0)，where 二is a general angle in the range 0 二一二。Plot the variation of the magnitude of thetraction vector Tn as a function of 日.3、利用应变协调条件检查其应变状态是否存在存在?i i | d 又环 b3% 讥3%(垃n»> ' =r 虽L* *dxdxtn dx idxjJxj3戈 idxkC1) x=Axy2, y=Bx2y, xy=0, A、B 为

4、常数2 2 2；x =k(x y ), ；y =ky , xy =2kxy k 为常数_x2y2 xz/c22(2) Zjj = y z ZI2匚I xz z 54、The displacements in an elastic material are given by2.M)(x22EI上),042M (1)M (1 J.xy,v 二El2EIwhere M, E， I, and l are constant parametersDetermine the corresponding strain and stress fields and show that this problem r

5、eprese nts the pure bending of a recta ngular beam in thex,y pla ne.5、写出如下问题的边界条件(a)用直角坐标，(b)用极坐标yx6、Express all boun dary con diti ons for each of the problems illustrated in the follow ing figure.也)(b)7、Show that the following stress components satisfy the equations of equilibrium ilh zero body fo

6、rces, bul are noi the solution io a problem iti elasticily:6 = ciy1 + 一 ”£)舟=+ 如 _ F)(Tj = rv(.v" -by2)隔=-2ivxyt>2 = ta = 0, c = constaiii 丰 0Sho that the Airy function <j)_?r4rfp1J1L+ 1 solves the folio cantile er 3beam problem, as siiov* n in the following figure. An ll>ujI fbi

7、such problems, boundlitjcoiidiliis al (tie ends (.1 0 and L should be foniiulaied only in terms of the reMiluintftm'e xystefrj, hiJe al y = 土f (he exact pointspecitication should be used, for (hecd?ie uith Ar 0, compaie the elasiicicy sliest field 議 ilhih弋 correspending insults fromstrength of m

8、aterials theory.The cantilever I team show n in【he figure is su tijecied lo a diMnbuied sheai stress :沁 7 on the upper face. The following Airy stress fijnetton is proposed for this problem0 = ciy 44 cjv4 + qv5 + cr + cr23i 4- “C +Determine the coiuaants and find (N? stre disidbution in the beant Us

9、e resultant force boundary conditions a( the ench. (Azlv: C Tltc/2L ci j/20人巧 -唧23J9、4正方形薄板三边固定，另一边承受法向压力p posi吟，如图所示，设位移函数为u =0jixnyv = a2 sin sin - b2b/（第4题图）OC!|TIb-|!a:ABy（第6题图）5、悬臂梁在自由端受集中力P作用，如图所示。试用极小势能原理求最大挠度第5题图提示设梁的挠曲线为w = a2x2 a3x36、对给定的应力函数：32223(1) Ax y , 2 = Bx y , 3 = Cxy，试确定它们哪个能作为平面

10、问题的应力函数,分析它们能解什么问题?3(2)证明=3Fxxy2r y2可以作为应力函数，并求在区域4c 3c 4c域内的应力分量，并分析该应力函数可以解决那类平面问题。x> 0, -eV, yY c 区二 Ax3 Bx2 ,7 如图所示矩形截面柱承受偏心载荷作用，且不计其重量，若应力函数 试求:u=0）,求位移分量(1)应力分量；(2)应变分量；(3)假设D点不移动，且该点处截面内线单元不能转动y P8、图示三角形截面梁只受重力作用，梁的质量密度为 求解各应力分梁。，宽度为1，试用纯三次应力函数f二二mm/9如图所示的楔形体两侧面上受有均布切向载荷q,试求其应力分量。yx10.已知一圆形薄管，平均半径为a,厚度为t,在薄管的两端受有拉力p和扭矩T作用，写出管内一点处的Tresca屈服条件和Mises屈服条件表达式。11 如图所示的矩形薄板 OABC , OA边与BC边为简支边，0C边与AB边为自由边。板不 受横向荷载，但在两个简支边上受大小相等而方向相反的均布弯矩M。试证，为了将薄板弯成柱面，即w=f（x），必须在自由边上施加以均布弯矩、.M。并求挠度和反力。12如图所示的矩形板，使用板的挠度表示