1、Mechanics of Materials112291 SUMMARY92 SKEW BENDING93 COMBINATION OF BENDING AND TORSION9-4 9-4 BENDING AND TENSION OR COMPRESSION ECCENTRIC TENSION OR COMPRESSION KERNEL OF THE SECTION3391 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4491 SUMMARYMPRzxyPP1 1、Composite def
2、ormation：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 deform
3、ation.55一、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MPRzxyPP66Phg g77Phg g88DamqPhg g99水坝水坝qPhg g1010 2 2、Methods to study composite deformation sPrinciple of superpositionAnalysis of external forces：External forces are reduced along the centroid of section and
4、 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，do the superposition of the stresses and establish the strength condition of the cr
5、itical point.1111二、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。121292 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、M
6、ethods 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 jxyzPPyPz131392 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
7、的平面弯曲。PyPzPzPyyzPj j14xyzPPyPz13142).Sum：Analyze bending in two perpendicular planes and sum the results of the calculation.xyzPyPzPPzPyyzPj j15152.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 j16jsinPPyjcosPPzSolution：1.Resolve the external force along the centroid principal axis of inertia of the c
8、ross section2.Study the bending in two planes:jjsinsin)()(MxLPxLPMyzjcosMMyxyzPyPzPPzPyyzPj jLmmx1717jsinPPyjcosPPz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 j18jcos yyyIMIzMz jsin zzzIMIyMy)sincos(jjzyIyIzM Stress due to My:Stress due to M z：Resultant stress：LPzPyyzPj jxyz
9、PyPzPLmmx1919jcos yyyIMIzMz jsin zzzIMIyMy )sincos(jjzyIyIzM My引起的应力：M z引起的应力：合应力：20LPzPyyzPj jxyzPyPzPLmmx200)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 t
10、he 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.21210)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
11、力点。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 22Example 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 axis2323 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242maxmax1maxDy
13、yzzDLWMWM中性轴中性轴ffzfy yzLxPyPzPhbPzPyyzPj jD2D1 24 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 secti
14、on and check the rigidity of the beam.N/m358447.0800sinqqySolution：Analysis of external forceresolve q yyzzWMWMmaxN/m715894.0800cosqqzNm40383358822maxLqMyzNm80483715822maxLqMzy 2634 hbyzqqLAB2525 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：外力分
15、析分解q yyzzWMWMmaxN/m715894.0800cosqqzNm40383358822maxLqMyzNm80483715822maxLqMzy 2634 hbyzqqLAB2626 93 COMBINATION OF COMBINATION OF BENDING AND TORSION80 P2zyxP1150200100ABCD2727 93 弯曲与扭转的组合弯曲与扭转的组合80 P2zyxP1150200100ABCD2828Solution：Reduce the external force to the centroid of the section and resolv
16、e itEstablish the strength condition of the rod shown in the figure.Bending and torsion80 P2zyxP1150200100ABCD150200100ABCDP1MxzxyP2yP2zMx2929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 P2zyxP1150200100ABCD30150200100ABCDP1MxzxyP2yP2zMx30Sum the bending moments and plot the diagram of the resultant bending mom
17、ent.)()()(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 component3131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）xMnMnMn(Nm
18、)xM (N m)XMmaxM(Nm)Mmaxx3232WMxBmax1PnBWM12231)2(2223134r2222max4PnWMWM1xB1B1xB1B33xB1B2MyMzMnMx1xB2xBM1B33画危险面应力分布图，找危险点WMxBmax1PnBWM12231)2(2建立强度条件2222max4PnWMWM1xB1B1xB1B34xB1B2MyMzMnMx1xB2xBM1B223134r34213232221421r1xB1B223WMMn2275.0WMMMnzy22275.0WMMMnzyr222475.0WMMMnzyr2223223134r2222max4PnWMWM
19、WMMMnzy222353536223WMMn2275.0WMMMnzy22275.0WMMMnzyr222475.0WMMMnzyr2223223134r2222max4PnWMWMWMMMnzy222351xB1B213232221421r36Analysis 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 probl
20、em 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.37WMMMnzyr2223WMMMnzyr222475.037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。WMMMnzyr2223WMMMnzyr2
21、22475.0弯扭组合问题的求解步骤：弯扭组合问题的求解步骤：3838 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 third s
22、trength.Analysis of external forces：Bending and torsion80 P2zyxP1150200100ABCD150200100ABCDP1MxzxyP2yP2zMxSolution:3939 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：4040Analysis of internal f
23、orces：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)71.3x71.25407.051205.540.64141内力分析：危险面内力为：Nm3.71maxMNm120nM)8.01(03.014.31203.71324322 MPa5.97安全M Z (N m)X(Nm)M
24、zxMy (N m)XMy(Nm)x（Nm）xMnMnMn(Nm)xM (N m)XMmaxM(Nm)71.3x71.25407.051205.540.642WMMnr22max34294 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 compression：Deformati
25、on of the rod due to simultaneous action of transversal and axial forces.434394 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。PR44PxyzPMyxyzPMyMz44APxPzzxMIyMzyyxMIzMyyyzzxIzMIyMAP2、Analysis of stress：45PMyMzPMZMyxyzz y45APxPzzxMIyMzyyxMIzMyyyzzxIzMIyMAP二、应力分析二、应力分析：4
26、6PMyMzPMZMyxyzz y46000yyzzxIzMIyMAP4、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),(PPyzPNeutral axis47000yyzzxI
27、zMIyMAP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),(PPyzP48yz5、Kernel of section in the problem of the eccentric tension、compression：ayaz012zyPiay012yzPiazAfter knowing ay and az,The action range
28、 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 section4949yz五、（偏心拉、压问题的）截面核心：五、（偏心拉、压问题的）截面核心：ayaz012zyPiay012yzPiaz已知 ay，az 后，压力作
29、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5050MPa75.82.02.0350000 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 the force P=350kN are shown i
30、n 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.P300200200P20020051MPa75.82.02.0350000 max2AP11max1zWMAPMPa7.113.02.06503503.02.03500002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
31、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.P300200200P20020052mm5102010100201020Cz235100101210010CyI4523mm1027.7252010122010Solution：Analysis of the internal force is shown in the figure.Centroid of the slot in the coordinates shown in the figureNm5001053PMPPSlot Example 5 A steel plate shown
32、 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?53PPMN2010020yzyC1053mm5102010100201020Cz235100101210010CyI4523mm1027.72520
33、10122010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Nm5001053PMPP54PPMN2010020yzyC1054PPMNycIzMANmaxmaxMPa8.1628.37125 Analysis of stress is shown in the figure.73631027.710555001080010100As the hole is moved to the middle of the plate)100(10mm9.631108.16210100263ma
34、xxNAmm8.36 sox2010020yzyC5555PPMNMPa8.1628.37125应力分析如图73631027.710555001080010100孔移至板中间时)100(10mm9.631108.16210100263maxxNAmm8.36 x2010020yzyC56ycIzMANmaxmax56MPa7.351.07000163nWTMPa37.6101.050432APSolution：For the composite deformation of tension and torsion,the stressed state at the critical point
35、 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 safe.2234r MPa7.717.35437.622AAPPTT5757MPa7.351.07000163nWTMPa37
36、.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。MPa7.717.35437.622AAPPTT582234r58 Chapter 9 Exercises 1.A circular shaft of steel is deformed under tension and torsion.Try to write out the strength conditions.If it shows the tension-torsion-bending comp
37、osite 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 section mn induced by the axial force P.Solution:3.A rectangular-section beam is shown in the figure.Know
38、ing 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?WMANmax222864/42/aPaaaPaP59 第九章第九章 练习题练习题 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、
39、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？WMANmax222864/42/aPaaaPaP60 Soluion:If the beam is changed to have circular sections,yyzzWMWMmaxmaxmaxMPa9605.0075.061021075.005.06105.12323MPaoWMMyz8.9265.25.1323222max2maxmax61 解：解：如改为圆截面如改为圆截面 yyzzWMWMmaxmaxmaxMPaoWMMyz8.9265.25.1323222max2maxmaxMPa9605.0075.061021075.005.06105.12323626364
