英汉双语材料力学9教学提纲.ppt

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1、英汉双语材料力学9CHAPTER 9 COMPOSITE DEFORMATIONSCHAPTER 9 COMPOSITE DEFORMATIONS91 SUMMARY92 SKEW BENDING93 COMBINATION OF BENDING AND TORSION9-4 9-4 BENDING AND TENSION OR COMPRESSION ECCENTRIC TENSION OR COMPRESSION KERNEL OF THE SECTION32第九章第九章第九章第九章 组合变形组合变形组合变形组合变形91 概述概述92 斜弯曲斜弯曲93 弯曲与扭转的组合弯曲与扭转的组合9-

2、4 9-4 拉拉(压压)弯组合弯组合 偏心拉（压）偏心拉（压）截面核心截面核心4391 SUMMARYMPRzxyPP1 1、Composite deformation：Structural members will produce several types of simple deformations when subjected to complex external loads.The stress corresponding to each simple deformation can not be neglected when the magnitude of each stres

3、s has the same order.This kind of deformation is called the composite deformation.54一、组合变形一、组合变形：在复杂外载作用下，构件的变形会包含几种简单变形，当几种变形所对应的应力属同一量级时，不能忽略之，这类构件的变形称为组合变形。91 概概 述述MPRzxyPP65Phg g76Phg g87DamqPhg g98水坝水坝qPhg g109 2 2、Methods to study composite deformation sPrinciple of superpositionAnalysis of ex

4、ternal forces：External forces are reduced along the centroid of section and resolved along principal axes of inertia.Analysis of internal forces：Determine the internal-force equation and its diagram corresponding to each external force component and the critical section.Analysis of stresses：Plot the

5、 distribution diagram of the stress Plot the distribution diagram of the stress in the critical sectionin the critical section，do the superposition of the stresses and establish the strength condition of the critical point.1110二、组合变形的研究方法二、组合变形的研究方法 叠加原理叠加原理外力分析：外力向形心(或弯心)简化并沿形心主惯性轴分解内力分析：求每个外力分量对应的

6、内力方程和内力图，确 定危险面。应力分析：应力分析：画危险面应力分布图，叠加，建立危险点的强 度条件。121192 SKEW BENDING 1 1、Skew bending：After bending deformation,the deflection curve and the external forces（transversal forces）of the rod are not in the same plane 2 2、Methods to study the skew bending：1).Resolve：Resolve the external load along two

7、centroid principal axes of inertia of the cross section and get two perpendicular planar bending.PzPyyzP xyzPPyPz131292 斜弯曲斜弯曲一、斜弯曲一、斜弯曲：杆件产生弯曲变形，但弯曲后，挠曲线与外力（横 向力）不共面。二、斜弯曲的研究方法二、斜弯曲的研究方法：1.分解：将外载沿横截面的两个形心主轴分解，于是得到两个正交的平面弯曲。PyPzPzPyyzP 14xyzPPyPz13132).Sum：Analyze bending in two perpendicular planes

8、 and sum the results of the calculation.xyzPyPzPPzPyyzP 15142.叠加：对两个平面弯曲进行研究；然后将计算结果叠加起来。xyzPyPzPPzPyyzP 15Solution：1.Resolve the external force along the centroid principal axis of inertia of the cross section2.Study the bending in two planes:Internal forcesInternal forcesxyzPyPzPPzPyyzP Lmmx1716解：

9、1.将外载沿横截面的形心主轴分解2.研究两个平面弯曲内内力力18PzPyyzP 17StressesStressesStress due to My:Stress due to M z：Resultant stress：LPzPyyzP xyzPyPzPLmmx1918应应力力My引起的应力：M z引起的应力：合应力：20LPzPyyzP xyzPyPzPLmmx19Maximum normal stressMaximum normal stressCalculation of the deformationCalculation of the deformationEquation of t

10、he neutral axisEquation of the neutral axisIt is obvious that only as Iy=Iz the neutral axis is perpendicular to the external force.As As =，it is the planar bending.it is the planar bending.PzPyyzP D1D2a aNeutral axisffzfy The maximum normal stress of tension or compression occurs in the points that

11、 lie in two sides of the neutral axis and have the farthest distance to the neutral axis.2120最大正应力最大正应力变形计算变形计算中性轴方程中性轴方程可见：只有当Iy=Iz时，中性轴与外力才垂直。在中性轴两侧，距中性轴最远的点为拉压最大正应力点。当当 =时，即为平面弯曲。时，即为平面弯曲。D1D2a a中性轴中性轴22PzPyyzP D1D2a affzfy 21Example 1 1 Force P is through the center of section and makes an angle

12、 the with axis z in the beam as shown in the figure.Determine the maximum stress and deflection of the beam.Maximum normal stressMaximum normal stressCalculation of deformationCalculation of deformationAs Iy=Iz the beam produce planar bending.Solution：Analysis of the critical point is shown in the f

13、igureffzfy yzLxPyPzPhbPzPyyzP D2D1a aNeutral axis2322 例例11结构如图，P过形心且与z轴成角，求此梁的最大应力与挠度。当Iy=Iz时，即发生平面弯曲。解：危险点分析如图24最大正应力最大正应力最大正应力最大正应力变形计算变形计算变形计算变形计算中性轴中性轴ffzfy yzLxPyPzPhbPzPyyzP D2D1a a23 Example 2 A wood purline is shown in the figure.Its span is L=3m and a uniformly distributed load q=800N/m is

14、acting on it.The permissible stress and deflection of it is respectively =12MPa and L/200，E=9GPa，Try to determine the dimension of the section and check the rigidity of the beam.Solution：Analysis of external forceresolve qa a=2634hbyzqqLAB2524 例例2 矩形截面木檩条如图，跨长L=3m,受集度为q=800N/m的均布力作用，=12MPa，许可挠度为：L/2

15、00，E=9GPa，试选择截面尺寸并校核刚度。解：外力分析分解qa a=2634hbyzqqLAB2625 93 COMBINATION OF COMBINATION OF BENDING AND TORSION80 P2zyxP1150200100ABCD2726 93 弯曲与扭转的组合弯曲与扭转的组合80 P2zyxP1150200100ABCD2827Solution：Reduce the external force to the centroid of the section and resolve itEstablish the strength condition of the

16、rod shown in the figure.Bending and torsion80 P2zyxP1150200100ABCD150200100ABCDP1MxzxyP2yP2zMx2928解：外力向形心 简化并分解建立图示杆件的强度条件弯扭组合变形80 P2zyxP1150200100ABCD30150200100ABCDP1MxzxyP2yP2zMx29Sum the bending moments and plot the diagram of the resultant bending moment.Determine the critical Determine the cri

17、tical sectionsection(Nm)MzxMy(Nm)xMn(Nm)xM(Nm)MmaxxInternal equations and diagrams corresponding to each external force component3130每个外力分量对应的内力方程和内力图叠加弯矩，并画图确定危险面(Nm)MzxMy(Nm)xMn(Nm)xM(Nm)Mmaxx3231Plot the diagram of stress distributionPlot the diagram of stress distribution in the critical section

18、 and determine in the critical section and determine the critical point.the critical point.Establish the strength conditionEstablish the strength condition33xB1B2MyMzMnMxM32画危险面应力分布图，找危险点建立强度条件34xB1B2MyMzMnMxM333534363535Analysis of external forces：Reduce the external forces to the centroid of secti

19、on and resolve them.Analysis of stresses:Establish strength conditions.Steps of solving the problem of composite deformation of bending and torsion:Analysis of internal forces：Determine the internal equation and its diagram corresponding toeach external force component and critical section.3736外力分析：

20、外力向形心简化并分解。内力分析：每个外力分量对应的内力方程和内力图，确定危 险面。应力分析：应力分析：建立强度条件。弯扭组合问题的求解步骤：弯扭组合问题的求解步骤：3837 Example 3 A hollow circular shaft is shown in the figure.Its inside diameter is d=24mm and its outside diameter is The diameters of pulley B and D are respectively D=30mm D1=400mm and D2=600mm，P1=600N，=100MPa.Try

21、to check the strength of the shaft with the third strength.Analysis of external forces：Bending and torsion80 P2zyxP1150200100ABCD150200100ABCDP1MxzxyP2yP2zMxSolution:3938 例例3 图示空心圆轴，内径d=24mm，外径D=30mm，B 轮直径D 1 400mm，D轮直径 D 2600mm，P1=600N，=100MPa，试用第三强度理论校核此轴的强度。外力分析：弯扭组合变形80 P2zyxP1150200100ABCD15020

22、0100ABCDP1MxzxyP2yP2zMx解：4039Analysis of internal forces：Internal forces in the critical section are：Analysis of stressAnalysis of stress：It is safe(Nm)MzxMy(Nm)xMn(Nm)xM(Nm)71.3x71.25407.051205.540.64140内力分析：危险面内力为：应力分析：应力分析：安全(Nm)MzxMy(Nm)xMn(Nm)xM(Nm)71.3x71.25407.051205.540.6424194 BENDING AND T

23、ENSION OR COMPRESSION ECCENTRIC TENSION OR COMPRESSION KERNEL OF THE SECTION PRPxyzPMyxyzPMyMz1 1、Composite deformation of Composite deformation of bending and tension or compression：Deformation of the rod due to simultaneous action of transversal and axial forces.434294 拉拉(压压)弯组合弯组合 偏心拉（压）偏心拉（压）截面核

24、心截面核心一、拉一、拉(压压)弯组合变形：弯组合变形：杆件同时受横向力和轴向力的作用而产 生的变形。PR44PxyzPMyxyzPMyMz432、Analysis of stress：45PMyMzPMZMyxyzz y44二、应力分析二、应力分析：46PMyMzPMZMyxyzz y454、Critical point（Farthest point from the neutral axis）3、Equation of the neutral axisFor the problem of eccentric tension or compression47P（zP，yP）yzyzNeutra

25、l axis46四、危险点四、危险点（距中性轴最远的点）三、中性轴方程三、中性轴方程对于偏心拉压问题中性轴中性轴48P（zP，yP）yzyz47yz5、Kernel of section in the problem of the eccentric tension、compression：ayazAfter knowing ay and az,The action range of the compressive force.As the compressive force is acted in this range there are no tensile stresses in the

26、 section.May determine an action point of the force P.Neutral axisKernel of section4948yz五、（偏心拉、压问题的）截面核心：五、（偏心拉、压问题的）截面核心：ayaz已知 ay，az 后，压力作用区域。当压力作用在此区域内时，横截面上无拉应力。可求P力的一个作用点中性轴中性轴截面核心5049Solution：The stresses in the cross sections of the two poles are both compressive ones.Example 4 4 Two poles s

27、ubjected to the force P=350kN are shown in the figure.The section of one pole is unequal and the section of the other pole is equal.Try to determine the normal stress with maximum absolute value in the poles.MPPd51Fig.Fig.P300200200P20020050解：两柱横截面上的最大正应力均为压应力 例例4 4 图示不等截面与等截面柱，受力P=350kN，试分别求出两柱内的绝对

28、值最大正应力。图（1）图（2）MPPd52.P300200200P20020051Solution：Analysis of the internal force is shown in the figure.Centroid of the slot in the coordinates shown in the figurePPSlot Example 5 A steel plate shown in the figure is subjected to forces P=100kN.Try to determine the maximum normal stress；If the slot

29、is moved to the middle of the plate and the maximum normal stress is kept constant,how much should the width of the slot be?53PPMN2010020yzyC1052例例5 图示钢板受力P=100kN，试求最大正应力；若将缺口移至板宽的中央，且使最大正应力保持不变，则挖空宽度为多少？解：内力分析如图坐标如图，挖孔处的形心PP54PPMN2010020yzyC1053PPMN Analysis of stress is shown in the figure.As the

30、hole is moved to the middle of the plate2010020yzyC5554PPMN应力分析如图孔移至板中间时2010020yzyC5655Solution：For the composite deformation of tension and torsion,the stressed state at the critical point is shown in the figure.Example 6 A circular rod which the diameter d=0.1m is subjected to forces T=7kNm and P=

31、50kN as shown in the figure,=100MPa.Try to check the strength of the rod with the third strength theory.Therefore,the rod is safe.AAPPTT5756解：拉扭组合,危险点应力状态如图 例例6 直径为d=0.1m的圆杆受力如图,T=7kNm,P=50kN,=100MPa,试按第三强度理论校核此杆的强度。故，安全。AAPPTT5857 Chapter 9 Exercises 1.A circular shaft of steel is deformed under te

32、nsion and torsion.Try to write out the strength conditions.If it shows the tension-torsion-bending composite deformation,try to write out the strength conditions.2.The cross-section area of the square-section rod is reduced half at the section mn.Try to determine the maximum tensile stress at the se

33、ction mn induced by the axial force P.Solution:3.A rectangular-section beam is shown in the figure.Knowing b=50mm and h=75mm.Try to determine the maximum normal stress of the beam.If the beam is changed to have circular sections with the diameter d=65mm，what is the maximum normal stress?58 第九章第九章 练习

34、题练习题 一、钢圆轴为拉伸与扭转的组合变形，试写出一、钢圆轴为拉伸与扭转的组合变形，试写出其强度条件。若为拉伸、扭转和弯曲的组合变形，其强度条件。若为拉伸、扭转和弯曲的组合变形，试写出其强度条件。试写出其强度条件。二、方形截面杆的横截面面积在二、方形截面杆的横截面面积在 mn 处减少一处减少一半，试求由轴向载荷半，试求由轴向载荷 P 引起的引起的 mn 截面上的最大拉截面上的最大拉应力。应力。解：解：三、矩形截面梁如图。已知三、矩形截面梁如图。已知 b=50mm,h=75 mm，求梁内的最大正应力。如改为，求梁内的最大正应力。如改为 d=65mm 的的圆截面，最大正应力为多少？圆截面，最大正应力为多少？59 Soluion:If the beam is changed to have circular sections,60 解：解：如改为圆截面如改为圆截面 616263此课件下载可自行编辑修改，仅供参考！此课件下载可自行编辑修改，仅供参考！感谢您的支持，我们努力做得更好！谢谢感谢您的支持，我们努力做得更好！谢谢