1、1,Chapter 9 Torsion,2,第九章 扭 转,3,Chapter 9 Torsion,9-1 The Torsion of Equal Section Pole,9-2 The Torsion of Elliptic Section Pole,9-3 Membrane assimilation,9-4 The Torsion of Rectangular Section Pole,9-5 The Torsion of Ringent Thin Cliff Pole,Torsion,4,第九章 扭 转,9-1 等截面直杆的扭转,9-2 椭圆截面杆的扭转,9-3 薄膜比拟,9-4 矩
2、形截面杆的扭转,9-5 开口薄壁杆件的扭转,扭转,5,Material mechanics has solved the torsion problems of round section pole, but it cant be used to analyze the torsion problems of non-round section pole. For the torsion of any section pole, it is a relatively simple spatial problem. According to the characteristic of the p
3、roblem, this chapter first gives the differential functions and boundary conditions, which the stress function should satisfy of solving the torsion problems. Then , in order to solve the torsion problems of relatively complex section pole, we introduction the method of membrane assimilation.,Torsio
4、n,6,扭转,材料力学解决了圆截面直杆的扭转问题，但对非圆截面杆的扭转问题却无法分析。对于任意截面杆的扭转，这本是一个较简单的空间问题，根据问题的特点，本章首先给出了求解扭转问题的应力函数所应满足的微分方程和边界条件。其次，为了求解相对复杂截面杆的扭转问题，我们介绍了薄膜比拟方法。,7,9-1 The Torsion of Equal Section Pole,1. Stress Function,A equal section straight pole, ignoring the body force, is under the action of torsion M at its two
5、 end planes. Take one end as the xy plane,as shown in fig. The other stress components are zero except for the shear stress zx、zy,Substitute the stress components and body forces X=Y=Z=0 into the equations of equilibrium, we get,Torsion,8,扭转,9-1 等截面直杆的扭转,一 应力函数,设有等截面直杆，体力不计，在两端平面内受扭矩M作用。取杆的一端平面为 xy面
6、，图示。横截面上除了切应力zx、zy以外，其余的应力分量为零,将应力分量及体力X=Y=Z=0代入平衡方程，得,9,From the first two equations, we know,zx、zy are functions of only x and y, they have nothing to do with z. From the third formula:,Annotation : the differential equationsof equilibrium for spatial problems are:,According to the theory of diffe
7、rential equations, there must exist a function x,y, from it,The function x,y is called stress function of torsion problems.,a,Torsion,10,扭转,根据前两方程可见，zx、zy只是x和y的函数，与z无关，由第三式,注：空间问题平衡微分方程,根据微分方程理论，一定存在一个函数x,y，使得,函数x,y称为扭转问题的应力函数。,a,11,The note: when the body force is zero, the compatibility equations
8、in terms of stress components for spatial problems are,Substitute stress components into the compatibility equations which ignoring the body force, we can see: the first three formulas and the last are satisfied, the other two formulas demand,Torsion,12,扭转,注：体力为零时，空间问题应力分量表示的相容方程,将应力分量代入不计体力的相容方程，可见
9、：前三式及最后一式得到满足，其余二式要求,13,2. Boundary conditions,On the profiles of the pole, substitute n=0 and surface force components which are zero into the boundary conditions, we get that the first two formulas can always be satisfied, the third formula demands:,The note:the stress boundary conditions for spat
10、ial problems are:,Namely,Being at the boundary:,Torsion,14,扭转,二 边界条件,在杆的侧面上，将 n=0，及面力分量为零代入边界条件，可见前两式总能满足，而第三式要求,注：空间问题应力边界条件,15,Then we have,This illuminates that at the boundary of the cross-section, the stress function is a constant. Because the stress components dont change when the stress funct
11、ion subtracts a constant, we can suppose when it is a single successional section(solid pole):,c,At the arbitrary end of the pole, the shear stress composes torsion,Torsion,16,扭转,于是有,在杆的任一端，剪应力合成为扭矩,17,Integral step by step, and notice that equals to zero at the boundary,At last we get,d,Torsion,18,
12、扭转,分步积分，并注意在边界上为零,最后得到,d,19,3. Displacement Formula,According to the relations of stresses, strains and displacements, we get,After integral, we get,Torsion,20,扭转,三 位移公式,根据应力、应变、位移的关系可以得到,积分后得到,21,Where, K denotes the torsion angle per unit length. Ignoring the displacement of the rigid body, we get
13、:,Substitute them into the above first two formulas at the right,The above two formulas can be used to solve displacementcomponents w。.,e,f,Torsion,22,扭转,其中K表示杆的单位长度内的扭转角.不计刚体位移,代入前面右边前两式,上两式可用来求出位移分量w。,e,f,23,Differentiating the above two formulas with respect to y and x, then subtracting these two
14、, we get:,Obviously the above formula,Where C=-2GK.,Obviously, in order to seek the solution of the torsion problems, we only need to find the stress function . We make it satisfy the equations b, c and d，then we solve the stress components from formula a and give the value of the displacement compo
15、nents from formulas e and f.,Torsion,24,扭转,上两式分别对y和x求导，再相减，得,可见前面公式b中,的C=-2GK.,显然，为了求得扭转问题的解，只须寻出应力函数，使它满足方程b、 c和d，然后由a式求出应力分量，由式e 和f给出位移分量的值。,25,9-2 The Torsion of Elliptic Section Pole,The semi-major axes and semi-minor axes of the elliptic are a and b respectively, its boundary function is:,The s
16、tress function equals to zero at the boundary, so we fetch,Substitute it into,1,Torsion,26,扭转,9-2 椭圆截面杆的扭转,椭圆的半轴分别为a和b，其边界方程为,应力函数在边界上应等于零，故取,代入,1,27,We get,Then we have,Substitute it formula (1), we get,Form,Torsion,28,扭转,回代入1式得,由,29,We can get,Then we get,At last we have,Torsion,30,扭转,可得,于是得,最后得,3
17、1,We get the final solutions:,And from,The total shear stress at any point of the cross-section is,Torsion,32,扭转,最后得到解答,于是由,横截面上任意一点的合剪应力是,33,9-3 Membrane Assimilation,From the example of the last section, we know: for the simple equal section pole of elliptic, we just give the calculating expressio
18、n of shear stress at the cross-section, we havent pointed out the position and direction of the maximum shear stress at the section; but for the poles with not too complex section such as rectangular and thin cliff, it is considerably difficult to solve its precise solution, let alone the relatively
19、 complex section pole. For this reason, we introduce the method of membrane assimilation. This method is built at the basis of similative of the mathematic relation between the torsion problem of pole and elasticity membrane which is under the action of equal side pressure and exaggerates tight arou
20、nd.,Supposing there are even membrane, spreading it at the boundary which is equal to or proportionate to the section of the torsion pole. When under the action of small even pressure on the profile, the inner of the membrane will produce even tensility, each point on membrane will occur small uprig
21、htness angle change along z direction as shown in fig.,Torsion,34,扭转,9-3 薄膜比拟,由上节的例子可以看出，对于椭圆形这种简单等截面直杆，我们给出了横截面上剪应力的计算表达式，但却没有指出截面最大剪应力的位置及其方向；而对于矩形、薄壁杆件这些截面并不复杂的柱体，要求出其精确解都是相当困难的，更不用说较复杂截面的杆件了。为了解决较复杂截面杆件的扭转问题，特提出薄膜比拟法。该方法是建立在柱体扭转问题与受均匀侧压力而四周张紧的弹性薄膜之间数学关系相似的基础上。,设有一块均匀薄膜，张在与扭转杆件截面相同或成比例的边界上。当在侧面上受
22、着微小的均匀压力时，在薄膜内部将产生均匀的张力，薄膜的各点将发生图示 z 方向微小的垂度。,35,Fetch a small segment abcd of the membrane, as shown in fig. Its projection on the xy plane is a rectangle, which side lengths are dx and dy respectively. Suppose the pull of the membrane per unit width is T, then from the condition of equilibrium alo
23、ng z direction , we get:,After predigestion, we get,Torsion,36,扭转,取薄膜的一个微小部分abcd图示，它在xy面上的投影是一个矩形，矩形的边长分别是dx和dy。设薄膜单位宽度上的拉力为 T，则由z方向的平衡条件 得,简化后得,37,Namely,Moreover, obviously the uprightness angle of the membrane at the boundary equals to zero, namely,For q/T is a constant, the above two formulas ca
24、n be rewritten as,a,And the differential equation and the boundary condition which the stress function satisfies are:,Torsion,38,扭转,即,此外，薄膜在边界上的垂度显然等于零，即,而应力函数所满足的微分方程和边界条件为,39,Comparing formulab with formulaa, we see that and are all determined by the same differential equation and boundary conditi
25、on, so they inevitably have the same solution. Then we have:,Torsion,40,扭转,将式 b与式 a对比，可见 与 决定于同样的微分方程和边界条件，因而必然具有相同的解答。于是有,41,Suppose the volume between membrane and the boundary plane is V, and we notice that,Then we have,Thereby we have,d,From,Moreover, we get,e,Torsion,42,扭转,43,Adjust the pressur
26、e q of which the membrane is under, and make the rights of formulas c, d, e equal to one, then we can gain some conclusions as follows:,Thus it can be seen, the maximum shear stress at cross-section of the elliptic section wringed pole exists at two end points of the semi-minor axes, its direction i
27、s parallel to the semi-major axes.,Torsion,44,扭转,调整薄膜所受的压力q，使得c、 d、 e三式等号的右边为1，则可得出如下结论：,1 扭杆的应力函数等于薄膜的垂度z。,2 扭杆所受的扭矩M等于该薄膜及其边界平面之间的体积的两倍。,3 扭杆横截面上某一点处的、沿任意方向的剪应力，就等于该薄膜在对应点处的、沿垂直方向的斜率。,由此可见，椭圆截面扭杆横截面上的最大剪应力发生在短轴的两端点处，方向平行于长轴。,45,9-4 The Torsion of Rectangular Section Pole,一 The Torsion of Narrow an
28、d Long Rectangular Section Pole,Suppose the side lengths of the rectangular section are a and b. If a is large than b(as shown in fig), we call it narrow and long rectangle. From the membrane assimilation, we deduce that the stress function almost doesnt change along with x at most cross-section, th
29、en we have,Torsion,46,扭转,9-4 矩形截面杆的扭转,一 狭长矩形截面杆的扭转,设矩形截面的边长为a和b (图示) 。若ab ，则称为狭长矩形。由薄膜比拟可以推断，应力函数在绝大部分横截面上几乎不随x变化，于是有,则,成为,47,The stress components are:,From the membrane assimilation, we know, the maximum shear stress exists at the long side of the rectangular section. Its direction is parallel to
30、x axis, and its value is,Torsion,48,扭转,应力分量为,由薄膜比拟可知，最大剪应力发生在矩形截面的长边上，方向平行于 x轴，其大小为,49,2. Pole with Rectangular Section,At the basis of the stress function for narrow and long rectangular section pole, we choose the stress function for any rectangular section pole as follow,Substitute into the diffe
31、rential function:,And make satisfy the boundary conditions:,Torsion,50,扭转,代入微分方程,并使满足边界条件,51,We get,Spread the right of the above formula into the progression of at the range of y-b/2,b/2, then compare the coefficient of both sides, we get:,Substitute Am into , we get the certain stress function:,To
32、rsion,52,扭转,得到,将上式右边在y-b/2,b/2区间展为 的级数，然后比较两边的系数，得,将Am代入，得确定的应力函数,53,From the membrane assimilation, we know, the maximum shear stress exists at midpoint of the long side of the rectangular section if ab,Where the wring angle k is obtained from,Torsion,54,扭转,由薄膜比拟可以推断，最大剪应力发生在矩形截面长边的中点若ab,其中扭角 k 由,5
33、5,Torsion,56,扭转,求得,57,9-5 The Torsion of Ringent Thin Cliff Pole,Actually we always face ringent thin cliff poles from engineer problems, such as angle iron, trough, I shaped iron and so on. The cross-sections of these thin cliff poles are always composed of narrow rectangle which has the equal widt
34、h. Whatever straight or bent, from membrane assimilation, we know, if only the narrow rectangle has the same length and width, then the torsion and the shear stress at the cross-section of two wringed pole are almost the same values.,Torsion,58,扭转,9-5 开口薄壁杆件的扭转,实际工程上经常遇到开口薄壁杆件，例如角钢、槽钢、工字钢等，这些薄壁件其横截面
35、大都是由等宽的狭矩形组成。无论是直的还是曲的，根据薄膜比拟，只要狭矩形具有相同的长度和宽度，则两个扭杆的扭矩及其横截面剪应力没有多大差别。,59,Supposing ai and bi denote the length and the width of the i narrow rectangle of the cross-section for the wringed pole, Mi denotes the torsion which the rectangular section is undergone, M denotes the torsion of all the cross-
36、section, I denotes the shear stress near the midpoint of the long side of the rectangle, k denotes the wrest angle of the wringed pole. From the result of the narrow rectangle, we get:,From the later formula, we get,Torsion,60,扭转,设 ai 及 bi 分别表示扭杆横截面的第 i 个狭矩形的长度和宽度，Mi表示该矩形截面上承受的扭矩，M表示整个横截面上的扭矩，i代表该矩形
37、长边中点附近的剪应力，k代表扭杆的扭角。则由狭矩形的结果，得,由后一式得,61,Also,So we have:,Consequently we have:,It is noticeable that: the shear stress of the midpoint of the long side of the narrow rectangle is considerably precise. However, because of the existence of stress concentration, the local shear stress maybe is more lar
38、ger than the mentioned at the joint of two narrow rectangle.,Torsion,62,扭转,而,故有,从而有,值得注意的是：由上述公式给出的狭矩形长边中点的剪应力已相当精确，然而，由于应力集中的存在，两个狭矩形的连接处，可能存在远大于此的局部剪应力。,63,Solution: substitute into the equations of compatibility,We get,Namely,Torsion,64,扭转,解： 将 代入相容方程,得,即,65,Thereupon,The wringed pole hasnt hole,
39、 obviously satisfies the boundary conditions on profile. From the boundary condition at the end of the pole:,We get,Consequently,Torsion,66,扭转,故,扭杆无孔洞， 显然满足侧面边界条件 ，由杆端部边界条件,得,从而得,67,The shear stresses,For equilateral triangle, from the membrane assimilation, we know:the maximum shear stress exists a
40、t,Namely,The twisty angle per unit length is:,Torsion,68,扭转,剪应力,单位长度上的扭角,69,Exercise 9.2 consider a pole with equal section. Its twisty angle per unit length is K, the shear elasticity module is . A function , where are real constants, is a function under affirmance. Solve the question: what is the
41、conditions must satisfy, then can be viewed as stress function of the torsion problems?,Substitute into the equation, we get,Torsion,70,扭转,习题9.2 等截面杆单位长度上的扭转角为K ，剪切弹性模量为 ，若函数 ，其中 为实常量， 为待定函数。试问： 满足什么条件时， 可作为扭转问题的应力函数？,答： 若题意所给函数能作为扭转问题的应力函数，则该函数必须满足相容方程,将 代入后得,此即为 需满足的条件。,71,The End,Torsion,72,扭转,结 束,
