(英汉双语)工程力学组合变形PPT精选文档

1、Mechanics of Materials1291 SUMMARY92 SKEW BENDING93 COMBINATION OF BENDING AND TORSION9-4 9-4 BENDING AND TENSION OR COMPRESSION ECCENTRIC TENSION OR COMPRESSION KERNEL OF THE SECTION391 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9-4 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491 SUMMARYMPRzxyPP1 1、Composite de

2、formation：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 stress has the same order .This kind of deformation is called the composite def

3、ormation .5一、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MPRzxyPP6Phg g7Phg g8DamqPhg g9水坝水坝qPhg g10 2 2、Methods to study composite deformation sPrinciple of superpositionAnalysis of external forces：External forces are reduced along the centroid of section and r

4、esolved 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，do the superposition of the stresses and establish the strength condition of the cri

5、tical point.11二、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内力方程和内力图，确 定危险面。画危险面应力分布图，叠加，建立危险点的强 度条件。1292 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、Met

6、hods to study the skew bending： 1).Resolve：Resolve the external load along two centroid principal axes of inertia of the cross section and get two perpendicular planar bending .PzPyyzPj jxyzPPyPz1392 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

7、平面弯曲。PyPzPzPyyzPj j14xyzPPyPz132).Sum：Analyze bending in two perpendicular planes and sum the results of the calculation.xyzPyPzPPzPyyzPj j152.叠加：对两个平面弯曲进行研究；然后将计算结果叠加起来。xyzPyPzPPzPyyzPj jjsinPPyjcosPPzSolution：1.Resolve the external force along the centroid principal axis of inertia of the cross se

8、ction2.Study the bending in two planes : jjsinsin)()(MxLPxLPMyzjcosMMyxyzPyPzPPzPyyzPj jLmmx17jsinPPyjcosPPz解：1.将外载沿横截面的形心主轴分解2.研究两个平面弯曲jjsin sin)( )(MxLPxLPMyzjcosMMy18PzPyyzPj jjcos yyyIMIzMz jsin zzzIMIyMy )sincos(jjzyIyIzM Stress due to My :Stress due to M z ：Resultant stress：LPzPyyzPj jxyzPyPzP

9、Lmmx19jcos yyyIMIzMz jsin zzzIMIyMy )sincos(jjzyIyIzM My引起的应力：M z引起的应力：合应力：20LPzPyyzPj jxyzPyPzPLmmx0)sincos(00jjzyIyIzMjctgtg00yzIIzyIt is obvious that only as Iy = Iz the neutral axis is perpendicular to the external force. 2maxDL1maxDy22zyfffzyfftgAs As j j = ，it is the planar bending. it is the

10、planar bending. PzPyyzPj jD1D2 Neutral axisffzfy The maximum normal stress of tension or compression occurs in the points that lie in two sides of the neutral axis and have the farthest distance to the neutral axis.210)sincos(00jjzyIyIzMjctgtg00yzIIzy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

11、。22zyfffzyfftg当当j j = 时，即为平面弯曲。时，即为平面弯曲。D1D2 中性轴中性轴222maxDL1maxDyPzPyyzPj jD1D2 ffzfy Example 1 1 Force P is through the center of section and makes an angle the j with axis z in the beam as shown in the figure. Determine the maximum stress and deflection of the beam.2maxmax1maxDyyzzDLWMWM232322)3()

12、3(yzzyzyEILPEILPfffjtgtgzyzyIIffAs Iy = Iz the beam produce planar bending.Solution：Analysis of the critical point is shown in the figureffzfy yzLxPyPzPhbPzPyyzPj jD2D1 Neutral axis23 例例11结构如图，P过形心且与z轴成j角，求此梁的最大应力与挠度。232322)3()3(yzzyzyEILPEILPfffjtgtgzyzyIIff当Iy = Iz时，即发生平面弯曲。解：危险点分析如图242maxmax1maxD

13、yyzzDLWMWM中性轴中性轴ffzfy yzLxPyPzPhbPzPyyzPj jD2D1 Example 2 A wood purline is shown in the figure. Its span is L=3m and a uniformly distributed load q=800N/m is 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 sect

14、ion and check the rigidity of the beam.N/m358447. 0800sinqqySolution：Analysis of external forceresolve q yyzzWMWMmaxN/m715894. 0800cosqqzNm40383358822maxLqMyzNm80483715822maxLqMzy 2634 hbyzqqLAB25 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/200 ，E=9GPa，试选择截面尺寸并校核刚度。N/m358447. 0800sinqqy解

15、：外力分析分解q yyzzWMWMmaxN/m715894. 0800cosqqzNm40383358822maxLqMyzNm80483715822maxLqMzy 2634 hbyzqqLAB26 93 COMBINATION OF COMBINATION OF BENDING AND TORSION80 P2zyxP1150200100ABCD27 93 弯曲与扭转的组合弯曲与扭转的组合80 P2zyxP1150200100ABCD28Solution：Reduce the external force to the centroid of the section and resolve

16、 itEstablish the strength condition of the rod shown in the figure.Bending and torsion80 P2zyxP1150200100ABCD150200100ABCDP1MxzxyP2yP2zMx29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 P2zyxP1150200100ABCD30150200100ABCDP1MxzxyP2yP2zMxSum the bending moments and plot the diagram of the resultant bending moment.

17、)()()(22xMxMxMzy)( ; )( ; )(xMxMxMnzyM Z (N m)X(Nm)MzxMy (N m)XMy(Nm)x （Nm）xMnMnMn(Nm)xM (N m)XMmaxM(Nm)MmaxxInternal equations and diagrams corresponding to each external force component31每个外力分量对应的内力方程和内力图叠加弯矩，并画图)()()(22xMxMxMzy确定危险面)( ; )( ; )(xMxMxMnzyM Z (N m)X(Nm)MzxMy (N m)XMy(Nm)x （Nm）xMnMnM

18、n(Nm)xM (N m)XMmaxM(Nm)Mmaxx32WMxBmax1PnBWM12231)2(2223134r2222max4PnWMWM1xB1B1xB1B33xB1B2MyMzMnMx1xB2xBM1B画危险面应力分布图，找危险点WMxBmax1PnBWM12231)2(2建立强度条件2222max4PnWMWM1xB1B1xB1B34xB1B2MyMzMnMx1xB2xBM1B223134r213232221421r1xB1B223WMMn2275. 0WMMMnzy22275. 0WMMMnzyr222475. 0WMMMnzyr2223223134r2222max4PnWMW

19、MWMMMnzy2223536223WMMn2275. 0WMMMnzy22275. 0WMMMnzyr222475. 0WMMMnzyr2223223134r2222max4PnWMWMWMMMnzy222351xB1B213232221421rAnalysis of external forces：Reduce the external forces to the centroid of section and resolve them. Analysis of stresses Establish strength conditions.Steps of solving the prob

20、lem 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.37WMMMnzyr2223WMMMnzyr222475. 0外力分析：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。建立强度条件。WMMMnzyr2223WMMMnzy

21、r222475. 0弯扭组合问题的求解步骤：弯扭组合问题的求解步骤：38 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 to check the strength of the shaft with the thi

22、rd strength .Analysis of external forces：Bending and torsion80 P2zyxP1150200100ABCD150200100ABCDP1MxzxyP2yP2zMxSolution:39 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 P2zyxP1150200100ABCD150200100ABCDP1MxzxyP2yP2zMx解：40Analysis of internal

23、forces：Internal forces in the critical section are：WMMnr22max3Nm3 .71maxMNm120nM)8 . 01 (03. 014. 31203 .71324322 MPa5 .97It is safeM Z (N m)X(Nm)MzxMy (N m)XMy(Nm)x （Nm）xMnMnMn(Nm)xM (N m)XMmaxM(Nm)71.3x71.25407.051205.540.641内力分析：危险面内力为：Nm3 .71maxMNm120nM)8 . 01 (03. 014. 31203 .71324322 MPa5 .97安

24、全M Z (N m)X(Nm)MzxMy (N m)XMy(Nm)x （Nm）xMnMnMn(Nm)xM (N m)XMmaxM(Nm)71.3x71.25407.051205.540.642WMMnr22max394 BENDING AND TENSION OR COMPRESSION ECCENTRIC TENSION OR COMPRESSION KERNEL OF THE SECTION PRPxyzPMyxyzPMyMz1 1、Composite deformation of Composite deformation of bending and tension or compre

25、ssion：Deformation of the rod due to simultaneous action of transversal and axial forces.4394 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。PR44PxyzPMyxyzPMyMzAPxPzzxMIyMzyyxMIzMyyyzzxIzMIyMAP2、Analysis of stress：45PMyMzPMZMyxyzz yAPxPzzxMIyMzyyxMIzMyyyzzxIzMI

26、yMAP二、应力分析二、应力分析：46PMyMzPMZMyxyzz y000yyzzxIzMIyMAP4、Critical point（Farthest point from the neutral axis）3、Equation of the neutral axisFor the problem of eccentric tension or compression0)1 (20202020yPzPyPzPizziyyAPAizPzAiyPyAPyyzzLWMWMAPmaxyyzzyWMWMAPmax012020 yPzPizziyy47P（zP， yP）yzyz),(PPyzPNeutr

27、al axis000yyzzxIzMIyMAP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0)1 (20202020yPzPyPzPizziyyAPAizPzAiyPyAP012020 yPzPizziyy中性轴中性轴48yyzzLWMWMAPmaxyyzzyWMWMAPmaxP（zP， yP）yzyz),(PPyzPyz5、Kernel of section in the problem of the eccentric tension、compression：ayaz012zyPiay012yzPiazAfter knowing ay and az

28、, The action range of the compressive force. As the compressive force is acted in this range there are no tensile stresses in the section.May determine an action point of the force P. ),(PPyz012020yPzPizziyyNeutral axis),(PPyzPKernel of section49yz五、（偏心拉、压问题的）截面核心：五、（偏心拉、压问题的）截面核心：ayaz012zyPiay012yz

29、Piaz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),(PPyz012020yPzPizziyy中性轴中性轴),(PPyzP截面核心50MPa75. 82 . 02 . 0350000 max2AP11max1zWMAPMPa7 .113 .02 .06503503 .02 .03500002Solution：The stresses in the cross sections of the two poles are both compressive ones. Example 4 4 Two poles subjected to

30、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 .P300200200P200200MPa75. 82 . 02 . 0350000 max2AP11max1zWMAPMPa7 .113 .02 .06503503 .02

31、 .03500002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值最大正应力。图（1）图（2）MPPd52.P300200200P200200mm5102010100201020Cz235100101210010CyI4523mm1027. 7252010122010Solution：Analysis of the internal force is shown in the figure. Centroid of the slot in the coordinates shown in the figureNm5001

32、053PMPPSlot 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 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 ?53PPMN2010020yzyC10mm510201010

33、0201020Cz235100101210010CyI4523mm1027. 7252010122010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Nm5001053PMPP54PPMN2010020yzyC10PPMNycIzMANmaxmaxMPa8 .1628 .37125 Analysis of stress is shown in the figure. 73631027.710555001080010100As the hole is moved to the middl

34、e of the plate)100(10mm9 .631108 .16210100263maxxNAmm8 .36 sox2010020yzyC55PPMNMPa8 .1628 .37125应力分析如图73631027.710555001080010100孔移至板中间时)100(10mm9 .631108 .16210100263maxxNAmm8 .36 x2010020yzyC56ycIzMANmaxmaxMPa7 .351 . 07000163nWTMPa37. 6101 . 050432APSolution：For the composite deformation of tensi

35、on 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=50kN as shown in the figure, =100MPa. Try to check the strength of the rod with the third strength theory.Therefore, the rod is saf

36、e.2234r MPa7 .717 .35437. 622AAPPTT57MPa7 .351 . 07000163nWTMPa37. 6101 . 050432AP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。 MPa7 .717 .35437. 622AAPPTT582234r59 Chapter 9 Exercises 1. A circular shaft of steel is deformed under tension and torsion. Try to write out the strength conditions. If it