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1、1,CHAPTER 6 DEFORMATION OF BEAMS DUE TO BENDING,2,第六章 弯曲变形,材料力学,3,64 Determine deflections and angles of rotation of the beam by the principle of superposition,65 Check the rigidity of the beam,CHAPTER 6 DEFORMATION IN BENDING,66 Strain energy of the beam in bending,67 Method to solve simple statica
2、lly indeterminate problems of the beam,68 Strain energy in bending beam,61 Summary62 Approximate differential equation of the deflecture curve of the beam and its integral63 Method of conjugate beam to determine the deflect and the angle of rotation of the beam,4,61 概述62 梁的挠曲线近似微分方程及其积分63 求梁的挠度与转角的共
3、轭梁法,64 按叠加原理求梁的挠度与转角,65 梁的刚度校核,第六章 弯曲变形,66 梁内的弯曲应变能,67 简单超静定梁的求解方法,68 梁内的弯曲应变能,5,6 SUMMARY,Study range：Calculation of the displacement of the straight beam with equal section in symmetric bending.Study object：Do rigidity check for the beam；Solve problems about statically indeterminate beams（compleme
4、ntary equations are supplied by the conditions of deformation of the beam),DEFORMATION OF BEAMS DUE TO BENDING,6,6 概 述,弯曲变形,研究范围：等直梁在对称弯曲时位移的计算。研究目的：对梁作刚度校核； 解超静定梁（变形几何条件提供补充方程）。,7,1).Deflection：The displacement of the centroid of a section in a direction perpendicular to the axis of the beam. It is
5、 designated by the letter v . It is positive if its direction is the same as f，or negative.,3、The relation between the angle of rotation and deflection curve：,1、Two basic quantities of to measure deformation of the beam,小变形,DEFORMATION OF BEAMS DUE TO BENDING,2). Angle of rotation：The angle by which
6、 cross section turns with respect to its original position about the neutral axis .it is designated by the letter . It is positive if the angle of rotation rotates in the clockwise direction,or negative.,2、deflection curve：The curve which the axis of the beam was transformed into after deformation i
7、s called deflection curve. Its equation is v =f (x),8,健身增肌 二次发育,WeiXin,TaoBao,9,1.挠度：横截面形心沿垂直于轴线方向的线位移。用v表示。 与 f 同向为正，反之为负。,2.转角：横截面绕其中性轴转动的角度。用 表示，顺时针转动为正，反之为负。,二、挠曲线：变形后，轴线变为光滑曲线，该曲线称为挠曲线。 其方程为： v =f (x),三、转角与挠曲线的关系：,弯曲变形,一、度量梁变形的两个基本位移量,小变形,10,6-2 APPROXIMATE DIFFERENTIAL EQUATION OF THE DEFLECTU
8、RE CURVE OF THE BEAM AND ITS INTEGRAL,1、Approximate differential equation of the deflection curve,Formula (2) is approximate differential equation of the deflection curve.,Small deformation,DEFORMATION OF BEAMS DUE TO BENDING,11,6-2 梁的挠曲线近似微分方程及其积分,一、挠曲线近似微分方程,式（2）就是挠曲线近似微分方程。,弯曲变形,小变形,12,For the st
9、raight beam with the same shape and equal section area, approximate differential equation of the deflection curve may be written as the following form：,2、Determine the equation of the deflection curve (elastic curve),1).integral of the differential equation,2).Boundary conditions of the displacement
10、,DEFORMATION OF BEAMS DUE TO BENDING,13,对于等截面直梁，挠曲线近似微分方程可写成如下形式：,二、求挠曲线方程（弹性曲线）,1.微分方程的积分,弯曲变形,2.位移边界条件,14,Discussion： Fit to the thinner and longer beam that made up from linear elastic material when its deformation is of planar bending and smaller. May be applied to determine the displacements of
11、 the beam with the same section shape and equal section area acting various loads. Integrate constants may be determined by the geometric conditions相容（boundary conditions、continuity conditions). Advantages：Range that it be applied is wide，may determine directly accuracy solution； Defects：Complicated
12、 calculation.,Displacement conditions at the supports：,Continuity conditions：,Sliding conditions：,DEFORMATION OF BEAMS DUE TO BENDING,15,讨论： 适用于小变形情况下、线弹性材料、细长构件的平面弯曲。 可应用于求解承受各种载荷的等截面或变截面梁的位移。 积分常数由挠曲线变形的几何相容条件（边界条件、连续条 件）确定。 优点：使用范围广，可以编程求出较精确的数值解； 缺点：计算比较繁琐。,支点位移条件：,连续条件：,光滑条件：,弯曲变形,16,Example 1
13、Determine the elastic curves 、maximum deflections and maximum angles of rotation of the following beams.,Set up the coordinates and write out the bending moment equation：,Write out the differential equation and integrate it,Determinate the integral constants by the boundary conditions,Solution：,17,例
14、1 求下列各等截面直梁的弹性曲线、最大挠度及最大转角。,建立坐标系并写出弯矩方程,写出微分方程的积分并积分,应用位移边界条件求积分常数,弯曲变形,解：,L,18,Write out the equation of the elastic curve and plot its curve,The maximum deflection and the maximum angle of rotation,DEFORMATION OF BEAMS DUE TO BENDING,19,写出弹性曲线方程并画出曲线,最大挠度及最大转角,弯曲变形,20,Solution：Set up the coordina
15、tes and write out the bending moment equation,DEFORMATION OF BEAMS DUE TO BENDING, Write out the differential equation and integrate it,21,解：建立坐标系并写出弯矩方程,写出微分方程的积分并积分,弯曲变形,22,Determine the integral constants by boundary conditions,DEFORMATION OF BEAMS DUE TO BENDING,23,应用位移边界条件求积分常数,弯曲变形,24,Write ou
16、t the equation of the elastic curve and plot its curve,Maximum deflection and the maximum angle of rotation,DEFORMATION OF BEAMS DUE TO BENDING,25,写出弹性曲线方程并画出曲线,最大挠度及最大转角,弯曲变形,26,6-3 METHOD OF CONJUGATE BEAM TO DETERMINE THE DEFLECT AND THE ANGLE OF ROTATION OF THE BEAM,1、Usage of the method：Determi
17、ne the deflection and the angle of rotation of designated point for the beam,2、Theory base of the method：Similar analogy:,Above two formulas are similar. By the method of analogy we can transform the differential equation, in form, into relation equation between the internal force and the external l
18、oad, further we can transform the problems to determine the deflection and the angle of rotation into the one of bending moment and shearing force.,DEFORMATION OF BEAMS DUE TO BENDING,Differential equation of the deflection curve of the beam,Relation between the bending moment of the beam and the ex
19、ternal load acting on the beam,27,6-3 求梁的挠度与转角的共轭梁法,一、方法的用途：求梁上指定点的挠度与转角。,二、方法的理论基础：相似比拟。,上二式形式相同，用类比法，将微分方程从形式上转化为外载与内力的关系方程。从而把求挠度与转角的问题转化为求弯矩与剪力的问题。,弯曲变形,28,3、Conjugate beam（real beam and imagine beam）：,The same coordinates,The same geometric section shape,Corresponding equation of the real beam：
20、,Integral of “force” differential equation of the imagine beam.,Corresponding equation of the imagine beam：,DEFORMATION OF BEAMS DUE TO BENDING,29,三、共轭梁（实梁与虚梁的关系）：,x轴指向及坐标原点完全相同。,几何形状完全相同。,实梁对应方程：,虚梁“力”微分方程的积分,弯曲变形,虚梁对应方程：,30,The quantities of 下脚标带“0”are all that of coordinate origin.,Integral of “d
21、isplacement” differential equation of the real beam,Set up the “force” boundary conditions of the imagine beam according to the “displacement” boundary conditions of the real beam.,DEFORMATION OF BEAMS DUE TO BENDING,31,下脚标带“0”的量均为坐标原点的量。,实梁“位移”微分方程的积分,依实梁的“位移”边界条件建立虚梁的“力”边界条件。,弯曲变形,32,中间铰支座A,中间铰支座A
22、,中间铰A,中间铰A,DEFORMATION OF BEAMS DUE TO BENDING,33,中间铰支座A,弯曲变形,中间铰支座A,中间铰A,中间铰A,34,Summary：Relations between the equal section real beam and the imagine beam as following:, Coordinate origin and direction of axis x are all completely same.,Geometric shapes are completely same.,Set up the “force” boun
23、dary conditions of the imagine beam according to the “displacement” boundary conditions of the real beam.,Determine “displacement” of the real beam according to “internal force” of the imagine beam.,c ：middle hinge middle hinge,DEFORMATION OF BEAMS DUE TO BENDING,Hinged supports,Hinged supports,35,总
24、结：等截面实梁与虚梁的关系如下：, x 轴指向及坐标原点完全相同。,几何形状完全相同。,依实梁的“位移”边界条件，建立虚梁的“力”边界条件。,依虚梁的“内力”，求实梁的“位移”。,弯曲变形,36,Solution： Set up coordinates and the imagine beam,Example 2 Determine the displacement at point B (deflection and angle of rotation of the following straight beams with the same shape section and equal
25、section area.,x,DEFORMATION OF BEAMS DUE TO BENDING,Determine the bending moment of the real beam in order to determine loads of the imagine beam, Determine the shearing force and the bending moment at the point B of the imagine beam in order to determine the deflection and the angle at the same poi
26、nt of the real beam.,37,解： 建立坐标和虚梁,例2 求下列等截面直梁B点的位移（挠度和转角）。,求虚梁B点的剪力和弯矩，以求实梁B点的转角和挠度,求实梁的弯矩方程 以确定虚梁荷载,弯曲变形,38,B点之矩,DEFORMATION OF BEAMS DUE TO BENDING,Area of of left side of the point B,The moment of the area of of left side of the point B about point B,39,求虚梁B点的剪力和弯矩，以求实梁B点的转角和挠度,弯曲变形,B点之矩,40,Solu
27、tion： Set up coordinates and the imagine beam,Determine the shearing force and bending moment of point B in the imagine beam.,Find out bending moment equation so as to determine loads of the imagine beam.,q,qa2,qa,A,B,C,D,qa2/2,x,M,qa2/2,qa2/2,3qa2/8,D,DEFORMATION OF BEAMS DUE TO BENDING,41,解： 建立坐标和
28、虚梁,求虚梁B点的剪力和弯矩,求实梁的弯矩方程以确定虚梁荷载,q,qa2,qa,A,B,C,D,qa2/2,x,M,qa2/2,qa2/2,3qa2/8,弯曲变形,D,42,Determine the shearing force and the bending moment at point B of the imagine beam.,How about that if the point C move to right or left？,A,B,C,a,a,a,D,qa2/2,3qa2/8,DEFORMATION OF BEAMS DUE TO BENDING,43,求虚梁B点的剪力和弯
29、矩,C点左右位移怎样？,弯曲变形,A,B,C,a,a,a,D,qa2/2,3qa2/8,44,Convert the change of section into the bending moment.,Geometric shape：the length is not change，moment of inertia changed into I0 .,Equation corresponding to the real beam：,Equation corresponding to the imagine beam：,4、Method of conjugate beam for chang
30、e section beam：,其它与等截面直梁完全相同。,DEFORMATION OF BEAMS DUE TO BENDING,45,将截面的变化折算到弯矩之中去。,几何形状：长度不变，惯性矩变为I0 。,实梁对应方程：,虚梁对应方程：,四、变截面直梁的共轭梁法：,其它与等截面直梁完全相同。,弯曲变形,46,Example 3 Determine displacement at point C of the following change section beam. Knowing：IDE =2IEB =2IAD .,Solution： Set up coordinates and th
31、e imagine beam.,a,a,P,0.5a,A,B,C,D,E,DEFORMATION OF BEAMS DUE TO BENDING,47,例3 求下列变截面直梁C点的位移，已知：IDE =2IEB =2IAD 。,解： 建立坐标和虚梁,弯曲变形,a,a,P,0.5a,A,B,C,D,E,48,a,a,P,0.5a,A,B,C,D,E,Determine the shearing force and the bending moment at point C of the imagine beam.,DEFORMATION OF BEAMS DUE TO BENDING,49,弯曲
32、变形,a,a,P,0.5a,A,B,C,D,E,求虚梁C点的剪力和弯矩,50,6-4 DETERMINE DEFLECTIONS AND ANGLES OF ROTATION OF THE BEAM BY THE PRINCIPLE OF SUPERPOSITION,1、Superposition of loads： 多个载荷同时作用于结构而引起的变形等于每个载荷 单独作用于结构而引起的变形的代数和。,2、Superposition of structural forms（rigidization method of segment by segment）：,51,6-4 按叠加原理求梁的挠度
33、与转角,一、载荷叠加：多个载荷同时作用于结构而引起的变形 等于每个载荷单独作用于结构而引起的变形的代数和。,二、结构形式叠加（逐段刚化法）：,弯曲变形,52,Solution、Resolve the loads as shown in the figure,q,P,P,=,+,A,A,A,B,B,B,C,a,a,DEFORMATION OF BEAMS DUE TO BENDING,Determine the deformations of the beam under the action of simple load by looking up the table.,Example 4 D
34、etermine The angle of rotation of point A and the deflection of point C by theorem of superposition.,53,例4 按叠加原理求A点转角和C点 挠度。,解、载荷分解如图,由梁的简单载荷变形表， 查简单载荷引起的变形。,弯曲变形,q,P,P,=,+,A,A,A,B,B,B,C,a,a,54,q,P,P,=,+,A,A,A,B,B,B,C,a,a,sum up,DEFORMATION OF BEAMS DUE TO BENDING,55,弯曲变形,q,P,P,=,+,A,A,A,B,B,B,C,a,a
35、,叠加,56,Example 5 Determine the deflection of point C by theorem of superposition.,Solution：Divide the load infinitely as shown in the figure.,Determine the deformations of the beam under the action of simple load by looking up the table.,Sum up,C,DEFORMATION OF BEAMS DUE TO BENDING,57,例5 按叠加原理求C点挠度。
36、,解：载荷无限分解如图,由梁的简单载荷变形表， 查简单载荷引起的变形。,叠加,弯曲变形,C,58,Example 6 Explanation of theorem of rigidization method of segment by segment）：,=,+,DEFORMATION OF BEAMS DUE TO BENDING,59,例6 结构形式叠加（逐段刚化法) 原理说明。,=,+,弯曲变形,60,6-5 RIGIDITY CKECK OF THE BEAM,1、Rigidity conditions of the beam,Where is called permissible
37、angle of rotation；f/L is called permissible ratio of the deflection and the span. In general we can do three kinds of calculations about the rigidity by this conditions：,、Check the rigidity：,(In civil engineering，first is the strength and second is rigidity except special cases.）,DEFORMATION OF BEAM
38、S DUE TO BENDING,For the civil engineering:,、Design dimension of the section；、Determine the permissible load。,61,6-5 梁的刚度校核,一、梁的刚度条件,其中称为许用转角；f/L称为许用挠跨比。通常依此条件进行如下三种刚度计算：,、校核刚度：,、设计截面尺寸；、设计载荷。,弯曲变形,(但：对于土建工程，强度常处于主要地位，刚度常处于从属地位。特殊构件例外）,62,Example 7 A circular hollow tube is shown in the following fi
39、gure. Its inside and outside diameter is separately：d=40mm、D=80mm. E=210GPa，f/L=0.00001 at point C, =0.001rad at point B. Try check the rigidity of the rod.,=,+,+,=,DEFORMATION OF BEAMS DUE TO BENDING,63,例7 下图为一空心圆杆，内外径分别为：d=40mm、D=80mm，杆的E=210GPa，工程规定C点的f/L=0.00001,B点的=0.001弧度,试核此杆的刚度。,=,+,+,=,弯曲变形
40、,64,=,+,+,图1,图2,图3,Solution：Transform the structure and determine the deformations under the actions of each simple load by looking up the table.,DEFORMATION OF BEAMS DUE TO BENDING,65,=,+,+,图1,图2,图3,解：结构变换，查表求简单 载荷变形。,弯曲变形,66,=,+,+,图1,图2,图3,Determine the deformation under the action of complex load
41、s by superposition,DEFORMATION OF BEAMS DUE TO BENDING,67,=,+,+,图1,图2,图3,弯曲变形,叠加求复杂载荷下的变形,68,Check the rigidity,DEFORMATION OF BEAMS DUE TO BENDING,(rad),69,校核刚度,弯曲变形,70,1、Calculation of strain energy in bending：,66 STRAIN ENERGY OF THE BEAM IN BENDING,Strain energy is equal to work of external forc
42、es.Dont consider strain energy of shear and neglect,DEFORMATION OF BEAMS DUE TO BENDING,71,一、弯曲应变能的计算：,66 梁内的弯曲应变能,弯曲变形,应变能等于外力功。不计剪切应变能并略去,72,Example 8 Determine the deflection at point C of the beam as shown in the figure by the energy method. The beam is of equal section and straight.,Solution： S
43、train energy is equal to work of external forces.,Applied the symmetry we get：,Thinking：If the load is distributed load do we applied this method to determine the displacement at the point C？,DEFORMATION OF BEAMS DUE TO BENDING,73,例8 用能量法求C点的挠度。梁为等截面直梁。,解：外力功等于应变能,在应用对称性，得：,思考：分布荷载时，可否用此法求C点位移？,弯曲变形
44、,74,2、 Impacting problems of the beam,1).Assumption：Impact bodies are rigid ones； Neglect the potential energy of weight and dynamic energy of the impacted bodies； Impact bodies dont rebound； Neglect the loss of energy of sound、light、heat and so on(energy is conservative).,DEFORMATION OF BEAMS DUE T
45、O BENDING,Before impact,75,二、 梁的冲击问题,1.假设：冲击物为钢体； 不计被冲击物的重力势能和动能； 冲击物不反弹； 不计声、光、热等能量损耗（能 量守恒）。,弯曲变形,76,DEFORMATION OF BEAMS DUE TO BENDING,After impact,77,弯曲变形,冲击前、后，能量守恒，所以：,78,3、Calculation of dynamic response：,Solution：Determine the static deflection of point C.,The dynamic response is equal to t
46、he product of the static response and the coefficient of dynamic load.,DEFORMATION OF BEAMS DUE TO BENDING,Example 9 A structure is shown in the figure. AB=DE=L，A、C are separately the middle points of AB and DE . Determine the dynamic stress of section C under the impact of weight mg.,79,三、动响应计算：,解：
47、求C点静挠度,动响应计算等于静响应计算与动荷系数之积.,例9 结构如图，AB=DE=L，A、C 分别为 AB 和 DE 的中点，求梁在重物 mg 的冲击下，C 面的动应力。,弯曲变形,80,Coefficient of dynamic load,Determine dynamic stress in the section C,DEFORMATION OF BEAMS DUE TO BENDING,81,动荷系数,求C面的动应力,弯曲变形,82,6-7 METHOD TO SOLVE SIMPLE STATICALLY INDETERMINATE PROBLEMS OF THE BEAM,1、
48、Treatment method：Combined compatibility equation of deformation、physical equation and equilibrium equations to determine the whole unknown forces.,Solution： Set up the primary beam,Determine the degree of statically indeterminacy. The structure left after the redundant constraints are substituted by
49、 the reactions corresponding the redundant constraintsprimary structure。,=,A,B,83,6-7 简单超静定梁的求解方法,1、处理方法：变形协调方程、物理方程与平衡方程相结合，求全部未知力。,解：建立静定基,确定超静定次数，用反力代替多余约束所得到的结构静定基。,=,弯曲变形,A,B,84,Geometric equationcompatibility equation of deformation,+,=,Physical equationrelation between deformation and the for
50、ces,complementary equation,Solve other problem（reaction、stress、deformation etc.）,DEFORMATION OF BEAMS DUE TO BENDING,85,几何方程变形协调方程,+,弯曲变形,=,物理方程变形与力的关系,补充方程,求解其它问题（反力、应力、 变形等）,86,Geometric equationcompatibility equation of deformation,Solution： Set up the primary beam,=,Example 10 Determine reaction
