[工学]材料力学课件第四版刘鸿文.ppt

材料力学（三） 刘鸿文,弯曲变形 应力和应变分析 强度理论,Deformation of bending,State of stress and strain, theories of strength,初参数法,x,q,y,p,m,Initial parameter method,factorial,(a),Y=0,A,B,p,L,写出梁的挠曲线方程,=0,P188(2),见,A,B,p,X=0 y=0,X=L y=0,X=0,例6-4,Deflection equation of beam,若遇到中间有集中力，集中力偶，变化的分布力时，在 后面加一个相应的项，称 马可利项,例6-5,A,p,p,B,a,a,C,求梁B点的挠度， 转角.,Concentrated force, concentrated moment, distributed force,见p198 6-3(b),A,p,p,B,a,a,C,例6-6,写出梁的挠曲线方程,此题在表上查不到 的。用初参数法 就很方便。,a,a,a,p,例6-7 求静不定梁的挠曲线方程,m,L,q,比静定梁变形小5倍,Deformation is less than 5 times in comparison with determinate beam.,二、积分法求解变截面梁的挠曲线方程,A,B,a,2a,3p,EJ,2EJ,c,p,2p,m,y,x1,x2,在C截面上,X1=a，x2=0。 分别代入连续条件,Use integral method to solve flexure equation of beam with varying cross section,习题：6-10（a）（d） 6-11（b）（c）,6-4 用叠加法求弯曲变形,一、载荷叠加：原理是小变形和满足虎克定理。 由前面可知载荷叠加。一个复杂的受力情况可分成几个简单 的受力之和。,每个简单的受力情况可写成,例6-8,A,B,a,3a,p,c,求中点的挠度,A,B,a,3a,c,A,B,a,3a,c,pa,pa,Use superposition method to solve flexure deformation,Superposition of load is based on small deformation and Hookes law.,B,a,3a,c,A,查p190（9）,查p189（5）,A,B,4a,pa,p,例,6-9 求中点C的挠度。,B,c,A,q,L,分析：本题有二种解法,Ask for flexure of midpoint C,B,c,A,L,B,c,A,L,q,qdx,dx,x,一、将qdx看成集中力作用在距原点为x 处。用p190（9）式。,二、利用对称性：,B,c,A,L,B,c,A,L,B,c,A,L,q,q,q,+,=,Utilize symmetric condition,B,c,L,q,+,若求中点C的挠度。它用对称性原理 可知为,A,B,c,L,A,q,q,p,求中点的挠度。 二边为反对称的。中点的挠度为零 本题就容易了。,二、变形叠加 用于变截面梁的计算,原理：,X面上转角为A面转角和AX相对转角,之和。若A面的转角为已知。若求得AX相对转角，则x面转角也 可求出。为了求出AX相对转角，可令,为A点固定端。AX相对转角要注意原梁上,EJ和弯矩都要相等。在计算中一般用等效力系代替。,Superposition of deformation, for the calculation of beam with varying cross section,principle,Rotation angle,Point A is a fixed end, it is required that EJ and moment should keep the same value on the beam, if relative rotation angle of AX is calculated.,Equivalent force system,例6-10,B,c,+,A,D,2EJ,EJ,p,求中点的挠度,C,D,B,p,l,l,m,P202 6-13,例6-11 等强度梁受力p的作用，变形读数f，梁长L，厚为t 证明：梁顶上任一点的应变 和挠度f满足：,p,L,t,证明：,习题：6-27 6-40 6-42,6-5 简单静不定梁,A,q,B,C,L,L1,AB梁EJ为常数，BC杆EA为常数。求B点的反力。 解：去BC杆得AB静定梁称为静定基。,q,Rb,B,A,L,静定基不是唯一的。也可解除A点 的刚臂，变成简支梁。,Simple situation of Statically indeterminate beam,Bending rigidity (EJ) of beam AB is a constant, tension rigidity of bar BC is also a constant. Determine the reaction on point B.,通常结构的刚度介于二者之间，因此B点反力不能超过这个范围,二、比较内力和变形,x,Reactions at point B can not exceed the range,Compare internal force and deformation,变形,6-6 提高弯曲刚度的一些措施,一、改善结构形式，减少弯矩的数值。 有效缩短跨度使挠度有很大的减少。 二、选择合理的截面形状。使惯性矩增加。,例6-12 板厚t，受p力作用，设横面上的应力分布均匀。导出 x面上的剪应力分布规律。,p,p,a,b,x,L,h,Methods to increase bending stiffness,To change the form of beam structure, such as to reduce the level of bending moment and to shorten the span of beam.,Select reasonable shape of cross section, so as to increase moment of inertial.,The thickness of shown plate is t under the action of force p, assuming uniformly distributed stress on cross section, ask to derive the distribution of shear stress on x cross section.,p,p,a,b,x,L,h,dx,dx,y,例6-13 p力作用于杆的A端，而AB 杆放在刚性平面上。若p0.5ql。 求A点的位移。,p,x,A,C,B,分析：C点的曲率半径为无穷远，它 的弯矩为零。C点可看成固定端。,P force acts on A end of the shown bar, bar AB lies on rigid plane. If p0.5ql, ask for the displacement of point A.,公式 应力 变形 拉伸 扭转 弯曲,第七章 应力和应变分析 强度理论,7-1 应力状态概念,一、工程上遇到的杆件受力形式是多种多样的。但总可以根 据三种基本变形受力特点，分解成几种简单形式。在各种基 本变形的内力图中找出危险截面。根据横面上的应力公式， 找出最大应力值。根据叠加原理再构成复杂应力受力情况。,二、梁的危险截面上，上、下表面的正应力最大，剪应力为零。 中性轴处的剪应力最大，正应力为零。根据强度公式可判断截 面上任意一点有正应力也有剪应力，它的强度问题还要研究它 的不同方位的应力。,研究一点且不同方位的应力情况，称为一点的应力状态。,Analysis of stress and strain, theory of strength,Stress state,Investigation of stress variation in different directions on a point is called as stress state of the point.,In engineering, various loading forms on a member will be encountered. The complicated loading action can be decomposed into several simple situations in terms of the three underlying deformations of bar.,研究方法：用微小六面体。1、它有一个平面是横面。横面上 的应力由公式可知。2、相互垂直面上剪应力相等。3、相互垂直面上正应力之和为常数。,Method of Investigation,Utilizing an infinitesimal cube element, 1. One of its surfaces is the cross section and stresses on the section is calculated using stress equation. 2. Shear stresses on mutually perpendicular planes equal to each other (Theorem of conjugate shear stress). 3. Summation of the normal stresses on the cube is constant which is called as the first invariant of stress.,六面体dx，dy，dz 0. 可认为六个面上应力分布均匀。,相互平行的平面上的应力相等。过该点的平行平面上的应力，由平衡方程可知，它们相等。六面体六个表面上的应力表示过该点的三个方位上的应力。,Stresses on mutually parallel planes equal to each other.,Stresses on six surfaces of unit cube indicate the stresses along three orientations passing through the point.,dx,dy,dz,单元体特点： 1、六面体的每一个面有一个正应力二个剪应 力。六个面上有18个应力。二个相互平行面的 应力相等。六个面上有9个应力。二个相互垂 直平面的剪应力相等，六个面上只有6个应力.,3、主单元体上，三个主应力都不为零。称为三 向应力状态二个主应力不为零，一个为零，称 为二向应力状态。一个主应力不为零，二个为 零。称为单向应力状态。三个主应力都为零的 称为无应力状态。,Principal plane,Principal stress,Principal unit cube,biaxial stress state,triaxial stress state,uniaxial stress state,If three principal stresses are all zero, the stress state is called as a state in no stress.,2、只有正应力而没有剪应力的平面，称为主平 面。主平面上的正应力称为主应力。三个相 互垂直的平面都是主平面的称为主单元体。,例7-1,A,1m,1m,160kN,120,200,M=80kNm,50,40kN,120kN,120,40,120,40,60,求A点的单元体,分析：作出内力图,A,To plot internal force diagrams,Stress state at point A,0.5,化工机械中常见薄壁圆筒( ),内部受压力p Thin walled vessel usually used in chemical engineering is subjected to internal pressure p.,7-2 二向和三向应力状态的实例,N,N,p,L,p,Examples of biaxial and triaxial stress states,7-3 二向应力状态分析-解析法,a,f,e,Analysis of biaxial stress states-Analytical method,Normal stress is positive as it in tensile stress and shear stress is positive as its direction is up on left side of unit cube, or is down on right side of unit cube,x is positive, buty is negative, is positive as N rotates counterclockwise from x axis.,讨论：1、角度为二倍。,反映二个平行平面上的应力完全相等。若斜面的夹角超过90可用小于90的处理。,Angle in trigonometry is twice the rotation angle of normal line, it reflects that stresses on two parallel planes are completely identical. If the angle of oblique exceeds to 90 degree, we can select an angle less than 90 degree.,2、正应力和剪应力为角度的函数。,（7-5）式有二个角，它们相差90。一个为最大主应力, 另一个为最小主应力。,在第二章中告诉我们，脆性材料破坏时由拉应力产生的。最大主应力方位决定了破坏的方向。,Normal stress and shear stress are functions of rotation angle.,Equation 7-5 can lead two angles, their difference is 90 degree, one represents the maximum stress,Another one represents the minimum stress.,In Ch. 2, we have found that failure of brittle materials results from tension stress, the orientation of maximum principle stress determines the direction of failure.,（3）相互垂直面的正应力之和为常数。,Sum of normal stresses on mutually perpendicular planes is a constant.,Difference between 0 and 1 is 45 degree.,The orientations of the maximum shear stresses is in 90 degree difference.,The first stress invariant,例7-2 分析铸铁试件受扭时破坏面的 方位。,T,T,T,分析：已知D，T。由剪应力公式 可知剪应力的大小。方向以左上 右下，因内力图为负的。其方向如 图示。,破坏面如图示。方向垂直最大拉应力。,Figure out the orientation of failure plane as cast iron rod is subjected to torsion.,Known diameter D and torque T, the shear stress can be calculated by shear stress equation.,Failure plane is shown in the figure, it is perpendicular to the maximum tension stress.,7-4 二向应力状态分析-图解法,Analysis of biaxial stress state-method of geometric solution,center,radius,Equation of circle,d,D,E,F,A,C,B,A1,B1,o,2,2,Build a Cartesian coordinate system using transverse and longitudinal coordinates and,Find the positions of points D and d.,Connect D and d, line Dd intersects transverse axis at point c.,Use point c as center of cycle, line cd as radius to plot a cycle,Start from cd and rotate an angle of 2 in counterclockwise, then get point c from circumference.,d,D,E,F,A,C,B,A1,B1,o,证明：,讨论：1、应力圆圆周上一点D。代表单元体X面上的 应力情况。,2、应力圆直径上的二点分别代表单元体两个正 交面上的应力情况，它们转向相同，圆心角 是2倍。A1点是最大正应力。它的大小为oC+ CD，方向是顺时针。和X面上的剪应力方向 相同。且当xy时它对应的圆心角900,D is a point on circumference of stress circle and represents stress state on x plane of unit cube,Central angle is double the rotation angle . Point A1 represents the maximum normal stress. Rotation angle is in clockwise direction.,If xy, its corresponding central angle is less than 90 degree.,B1点是最小正应力.它的大小是,Point B1 represents the minimum normal stress, its value is,d,D,E,F,A,C,B,A1,B1,o,2,2,它的方向为逆时针.且当xy时 它对应的圆心角900,3、脆性材料受力破坏时，破坏的方向 和最大正应力有关。破坏角的方向 有三种方法确定。 由方程式(7-5)中得到的二个角分别代入(7-3)式得到。此法麻烦。,The rotation direction corresponding to B1 is anticlockwise, if xy, its central angle is larger than 90 degree.,When brittle materials fail under external load, the fracture direction is related to the maximum stress.,There exist three methods to determine the fracture direction. The two angles derived from equation 7-5 are substituted into equation 7-3 to obtain the fracture direction. The method is tedious,根据上面分析可知:由X面上的剪 应力方向来决定。如果x是正的,则顺时针方向的为第一主应力。,Based on the previous analysis, the orientation of maximum principal stress can be determined by the direction of shear stress. If x is positive, the maximum principal stress is located on the clockwise direction.,单元体上没有剪应力，正应力都相等。应力 圆是横轴上的一点圆。它的任何方向的截面都 是主截面。应力相等。,单向应力状态。圆周必定通过坐标原点。,纯剪切。圆心和坐标原点重合。,Pure shear,Center of circle coincides with the origin of coordinate,For uniaxial stress state, circumference of stress circle must pass through the origin of coordinate.,If there exists no shear stress and all normal stresses are identical on a unit cube. Stress circle is a point circle on transverse axis. Any given direction is the principal direction and all normal stresses are identical.,例,7-3 求主应力及其截面位置,100MPa,50MPa,50MPa,86.6MPa,86.6MPa,60,60,分析：x坐标可任意选。,60,30,Find the principal stresses and their orientations of cross section,例7-4 有一处平面应力状态的单元体, =30, =81.7MPa, =-18.3MPa; 面上， =200MPa, =50MPa, 求主应力 和主平面方位及最大剪应力。,解：圆半径相同。用下面公式,There is a unit cube in planar stress state. Find principal normal stresses and orientation of principal plane, as well as the maximum shear stress,习题：7-2 7-3（b）（d） 7-4（c）（e） 7-7,7-5 三向应力状态,x,y,z,N,px,pz,py,当受力杆件一点处三个主应力,分别沿x, y, z方向已知。,其上任意一个,斜面，它的外法线N和三个坐标轴夹角,A,C,B,Triaxial stress state,Three principal stresses at a point on stressed member,Resolve p along the normal line of oblique,tetrahedron,Included angles,x,y,z,N,px,pz,py,A,C,B,project them on the normal line of the oblique plane.,可以分别以 为圆心,Use the coordinates as centers of separate stress cycles, respectively.,习题：7-18(b,c) 7-19(c),In biaxial stress state,7-6 位移与应变分量,平面应变状态：杆件各点的位移和应变都发生在同一平面内。 在实际问题中，最大的应变通常发生在表面。 而表面的应变很容易求出。,M,N,L,M,L,N,dx,dy,u,v,Components of displacement and strain,Plane strain state: for any given point on a member, displacements and strains occur on the same plane. In practical problems, the maximum strain usually occurs on the surface of the member and it is easy to obtain the surface strain.,x,o,y,7-8 广义虎克定律,1、在单向受拉（压）时，若材料处于弹性阶段。我们有公式,在剪切时，我们有下面公式,2、定理：在小变形中，材料的线应变只能产生正应力。 剪应力只能产生剪应变。,证明：用反证法 若在正应力作用下，材料发生 剪应变。它绕x轴转动180。原来向外 变形现在向里。在同一力作用下，不 可能有二种变形形式。 同理可证明剪应力只能产生剪应变,generalized Hookes law,theorem,Under the condition of small deformation, line strain of materials just results in normal stress, while shear strain corresponds to shear stress.,Under the condition of uniaxial tension, materials deform elastically, we have the equation,For shearing deformation,verification,apagoge,Assume that under the action of normal stress, shear strain occurs in unit cube as shown in the figure. Turn the unit cube around x axis for 180 degree, then the original outward deformation is changed to inward deformation. It is impossible to exist different deformations under the action of same force.,3、在小变形情况下，线应变可由各个方向的正应力叠加而成,+,+,=,Under the condition of small deformation, line strain results from the superposition of normal stresses oriented from other orthogonal directions.,4、若单元体由三个主应力产生。,5、应力通常由应变的测量获得。在平面应变状态分析中， 可导出和应力相同的式子,If unit cube is composed of three principal stresses, then the principal strains equal to,Page 233, Eq. (7.13, 7.14),In the analysis of plane strain state, the same equations of strain can be derived as the stress.,6、体积改变量,volume change,Three-element 45 rosette can measure the line strains along 0, 45 and 90, substitute the strains into the above equations, we have,Plane Hookes law,Page 235,体积改变量只能和平均主应力有关。而和各个主应力的比值无关 只要单元体的平均主应力相同。它们的体积改变量相同。,The change in volume is just related to the average principal stress and has no relation with the ratio of each principal stress. Therefore if the average principal stresses (mean stress) is identical, the change in volume is the same.,Bulk modulus,Mean stress,例7-5 刚性槽内放入一立方体（长为1cm）的铝块。当受压力 p=6kN作用时。设它受力均匀。E=70GPa。 =0.33。 求它的三个主应力的大小和相应 的应变。,y,z,x,分析：x方向没有力,但有变形. y方向应变为零。但有力存在 Z方向受力均匀,有力又有变形,p,bulk modulus,Volumetric strain,Materials exceeds elastic deformation range,An aluminum cube with size 1cm is placed into a rigid slot, a force 6KN is applied to the cube. Assume the cube is stressed uniformly. Ask for the values of the three principal stresses and the corresponding strains.,There are no deformation restriction in x direction and no deformation in y direction.,y,z,x,p,x,z,y,p