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1、1Chapter 5:Internal Forces of Beams梁的内力1 Introduction2 Shearing force and bending moment equationsand diagrams 剪力、弯矩方程与图3 Relation between the intensity of loading,the shearing force,and bending moment in a straight beam载荷与剪力、弯矩间的微分关系4 Internal forces of frame and curved bar 刚架与曲杆的内力1 Introduction外力
2、或外力偶的矢量垂直于杆轴。Beams are designed to carry loads perpendicular to their longitudinal axis Bending of beamConcentrated force and moment;distributed load 力或力矩矢量垂直于轴线forces acting transverse to the axis of bar;couples having their plane of action parallel to the axis of bar.Deformation of beam:External F
3、orces and Supports of Beams轴线变弯The axis of barbend 通常以轴线代表梁2典型约束形式的反力Reactions at various supports约束类型Types of supportszPin support(固定铰支座)：Two forces acting at the pinFRx,FRyzRoller supports(可动铰支座)：A force normal to the beamFRzFixed support(固定端)：FRx,FRy and M梁的类型Types of beams常见静定梁Common types of st
4、atically determinate beamszSimply-supported beams:with a pin support at one end and a roller support at the other end简支梁：一端固定铰支、另一端可动铰支的梁zBeam with an overhang,or two overhangs 外伸梁：具有一个或两个外伸部分的简支梁zCantilever:fixed at one end and free at the other end悬臂梁：一端固定、另一端自由的梁静不定梁Statically indeterminate beams
5、The number of force quantities is bigger than the number of static equilibrium equations.（Chapter 6）拉压杆横截面上的内力轴力内力分析应力分析变形分析受扭轴横截面上的内力扭矩受弯曲梁横截面上的内力？2 Shearing force and bending moment equations and diagrams 剪力、弯矩方程与图剪力与弯矩Shearing force and bending momentmoment 剪力、弯矩方程与图Shearing force and bending mom
6、ent equations and diagramsExamplesShearing force and bending momentFS剪力梁的内力Internal force(stress resultants)in beams:Shearing force 剪力作用线位于所切横截面的内力Bending moment一侧)(1S=niiFFM弯矩=0 0S1FFF,FAyy1S FFFAy=+=0-)(01bFabFM,MAyC)(1abFbFMAy=一侧)(1=niCimMBending moment 弯矩矢量位于所切横截面的内力偶矩剪力与弯矩的符号规定Sign convention f
7、or Fs、Mpositive when act clockwise about the surface on which it acts(使微段沿顺时针方向转动的剪力为正)positive when compress the upper part of the beam使横截面顶部受压的弯矩为正；使微段弯曲呈凹形的弯矩为正3Calculation of shearing force and bending moment 剪力与弯矩计算nFind the reactions求反力oAn imaginary cutting plane is passed through the beam at
8、the position of interest,one half of the beam can then be separated.The internal forces FS and moments M are are assumed to be positive.假想地将梁切开，并任选一段为研究对象画所选梁段的受力图M为研究对象，画所选梁段的受力图，FS 与 M 宜均设为正p Fy=0 FSq MC=0 M，C：the centroid of cross section截面形心剪力、弯矩方程与图Shearing force and bending moment equations an
9、d diagrams)(SSxFF=)(xMM=Fs and M equations 剪力、弯矩方程2qlFFByAy=qxFFAy+=S)(0lx 2xqxxFMAy+=)(0lx qxqlF+=2S222xqxqlM+=Shearing force and bending moment diagrams剪力、弯矩图qlqlstraight line)(0 2SlxqxqlF+=z表示 FS与 M 沿杆轴（x轴）变化情况的图线，分别称为剪力图与弯矩图2)(2-(0)SSlF,F=parabola 二次抛物线)(0 222lxxqxqlM+=z2maxS,qlF=82maxqlM=z Exam
10、pleExample 5-2 Construct diagrams of Fsand MSolution：1.Reactions2.Equilibrium equations)(0 ,1S1axlbFFFAy=)(0 ,2S2bxlaFFFBy=4.DiscussionAt the point of application of a concentrated force F,Fs changes abruptly by an amount equal to F在 F 作用处,左右横截面上的剪力值突变,弯矩连续FFF=左右SS2qaExample 5-3 Construct diagrams o
11、f Fsand MSolution：1.Reactions2 M,qaFCCy=2.Equilibrium equationssection AB:section BC:)(0 11S1axqxF=)(0 2S2axqaF=)(0 21211axqxM=)(0 22222axqaqaxM=43.Shearing force and bending moment diagrams1S1qxF=qaF=S2211qxM=222qaqaxM=no4.DiscussionAt the point of application of a concentrated moment Me,M changes
12、abruptly by an amount equal to Me 在 Me 作用处,左右横截面的剪力连续,弯矩值突变eMMM=左右22p22maxqaM=maxSqaF=BAaRACq3aRBAx1FS1M1Cx2FS2M2qaRqaRBA3834=11134qxqaqxRFAS=2112111213421qxqaxqxxRMA=1.支反力2.剪力与弯矩RA()axqxFS=2220()axqxM=2222021982maxqaM=ax341=剪力、弯矩方程与剪力、弯矩图总结：4分析步骤：求支反力建立坐标建立剪力弯矩方程画剪力弯矩图4当需要分段列剪力弯矩方程时（集中力，集中力偶，分布载荷不作
13、用在整个梁段），可以采用整体坐标或局部坐标。4剪力弯矩图中要标明符号以及特征点的大小。4剪力图在有集中力作用处（包括支座处），发生突变。4弯矩图在有集中力偶作用处（包括支座处），发生突变。Example 5-5 The load can move along the beamTo find:Maximum value of FSand M1.FSand M diagramslFlFAy)(=lFlFFAy)()(S=lFFMAy1)(FFF=)0(SmaxS,4)2(maxFllMM=lFM21d)(d0=2l=2.Maximum value of FS and MSolution：1.Rea
14、ctionsExample Construct diagrams of Fsand M 非均布载荷作用下，画剪力与弯矩图图2oRlqF=6olqFAy=3olqFBy=2.Shear force and bending moment equationslxqxlqF00S26=32600 xlxqxxlqM=200S26xlqlqF=30066xlqxlqM=200S26xlqlqF=3qlq3.Shear force and bending moment diagrams parabola0066xlxM=cubicq0l2/(9 3)026200S=xlqlqF3/lxC=39662030
15、0lqxlqxlqMCCC=53载荷集度、剪力与弯矩间的微分关系Relations between the intensity of loading,the shearing force,and bending moment in a straight beam载荷集度、剪力与弯矩间的微分关系Differential Equations about Load,Shear force,and Bending MomentShear force,and Bending Moment 利用微分关系画剪力与弯矩图Shear and Moment Diagrams by Inspection Examp
16、les载荷集度、剪力与弯矩间的微分关系Differential Equations about Load,Shear force and Bending Moment(a)0)d(d 0SSS=+=FFxqF,Fyd(b)0d2dd 0S=+=MxFxxqMM,MCqxF=ddSSddFxM=qxM=22ddq:assumed positive when act upward 向上为正;x:measured from the left end to right向右为正.0The conditions for above relationships to be valid：利用微分关系画剪力与弯
17、矩图Shear and Moment Diagrams by InspectionMeaning of the relationships:qxF=ddSSddFxM=qxM=22ddMechanics:equilibrium equations for an element Geometry:Application:Shear and Moment Diagrams by Inspection4剪力图某点处的切线斜率=该截面处载荷集度的大小the slope of Fsis equal to q4弯矩图某点处的切线斜率=该截面处剪力的大小The slope of M is equal to
18、Fs4该截面处载荷集度的正负决定弯矩图某点处的凹凸性Concave/convex of Mis determined by positive/negative of qqxF=ddSSddFxM=qxM=22ddq(x)=0q(x)=c 0qUniform load distribution斜线2次凸曲线2次凹曲线M图Fs图qxF=ddSSddFxM=qxM=22ddqq(x)=ax+b(a 0)q(x)=ax+b 0(a 0)q(x)=ax+b 0(a 0)M图Fs图2次凹曲线2次凸曲线2次凹曲线2次凸曲线3次凸曲线3次凸曲线3次凹曲线3次凹曲线)At the point of applic
19、ation of a concentrated force F,Fs changes abruptly by an amount equal to F（在 F 作用左右横截面上的剪力值突变F）M右qF左M左F右dxFF左+q(x)dx+F=F右M左+F左dx+Fdx/2+q(x)dx2/2=M右F左+F=F右M左=M右0F向上为正)At the point of application of a concentrated moment Me,M changes abruptly by an amount equal to Me（在Me作用处左右横截面的弯矩值突变Me）F左+q(x)dx=F右M
20、左+F左dx+M+q(x)dx2/2=M右F左=F右M左+M=M右qF左M左F右M右dxM0M顺时针为正6lMFFByAye=利用微分关系画剪力弯矩图的步骤：1.Find reactions at the ends求支反力2.Determine the values for F and M at the ends ofDetermine the values for Fsand M at the ends of each region 求各段曲线端点的剪力、弯矩值lMFFAyAeS)(=+ee0MFMMAyA=+00=ByBFM3.Identify regions 确定各段曲线的形状0切记：q
21、向上为正；x向右为正。nq=0，FS horizontal line，M straight lineoTo find(FS)A+FS diagrampTo find(M)A+and(M)B M diagramlMFAeS-)(=+eMMA=+0=BM4.Fsdiagram 5.M diagram FS horizontal lineM slope lineThe equationscan be integratedto thatat any spanwiselocation,the changes of bending momentequal to the areaunder the shea
22、r diagramqxF=ddSSddFxM=x2面积法(积分法)：dxFMMxs=112anddxqFFxxSS=2112总之，Fs:跟着F箭头走，段内变化看q面积M:Me顺时针向上走，段内变化看Fs面积从零开始，回到零 ExamplesConstruct diagrams of Fsand MSolution：斜线ql/80ql2/16ql/8-3 ql/8ql2/1602.Identify regions 1.Values of Fsand M 3=Dlx3.Fs and M diagram 1 2 Dx83lxD=1289 832838322qllqlqlMD=9ql2/1287 组合
23、梁的内力分析：2FaFFF1F22FaF1F2Example Construct diagrams of Fsand MSolution：1.AnalysisnSeparate the combined beamso Concentrated force and moment:FSand M diagrams are constructed by straight lines2FFFCyAy=23FFDy=23FaMD=2.Reactions3.FSdiagramhorizontal line423maxSFF=A pin can transmit force,not moment(铰链传力不
24、传力偶矩).At the two sections combinedwith the gemel,M=0.4.M diagramstraight line23maxFaM=Solution：1.ReactionsolqF=olqF=olqF=Example Construct diagrams of Fsand M2R6Ay3By200S26xlqlqF=30066xlqxlqM=2.Shear force and bending moment diagrams parabola cubicq0l2/(9 3)026200S=xlqlqF3/lxC=396620300lqxlqxlqMCCC=
25、4.Check-up FS and M diagramsFS parabolaM cubicqFddd2S2=obaxq+=nxxdq 0FSconvex curve0dd22=qxMpM convex curve When FS=0 Maximum value of M；When q=0 Maximum value of FS.n利用微分关系，确定各梁段剪力、弯矩图的形状o计算各梁段起、终与极值点等控制截面的剪力与弯矩将上述二者结合绘制梁的剪力 微分关系法要点p将上述二者结合，绘制梁的剪力、弯矩 图q在集中载荷作用下，梁的剪力与弯矩图一定由直线所构成在分布载荷作用下，剪力或弯矩图线的凹凸性，
26、由其二阶导数的正负确定84 Internal forces of rigid frame and curved bar刚架与曲杆的内力 Frame Curved barformed by bars connected by rigid joints（刚性接头）Framerigid joints：zRestrict the relative displacement and rotation约束相对线位移与角位移zTransmit both force and moment可传力，也可传递力偶矩Internal forces of frame1.reactions=0 ,0 ,0 yAxFMF/
27、2 ,qaFFqaFAyCyAx=2.Equations of internal forces qBC：,2S1qaF=AB：,2S2qxF=2N2qaF=112xqaM=2222-2xqaqaM=3.Diagrams of internal forces 2S2S1 ,2qxFqaF=2 ,0N2N1qaFF=222112-2 ,2xqaqaMxqaM=No moment at rigid joints:M is continuous如刚性接头处无外力偶，则弯矩连续M diagram:paint at the compressedside of the section when bendin
28、g画在横截面弯曲时受压一侧FNFQFQFNFNFQFNFQMMMMCurved barBar with curved axis before the application of load 未受力时，轴线即为曲线的杆件9sinSFF=cosNFF=n Components of the Internal Forces：轴力，剪力，弯矩Internal forces of curved bar 曲杆内力)cos-(1FRM=Mpositive when act to increase curvature 使杆微段愈弯的弯矩为正FS，FNsame with beforeoSign conventi
29、onpaint at the compressedside of the section when bending 画在截面受压的一侧。pBending moment diagram一般建立极坐标来表示截面的位置 组合刚架的内力分析：qCa/2aF2N2qCa/2aa/2aqaa/2aF1N1FAyFAxAMAABqaFByBqaN1N2F1F2qaN1=qa F1=FBy=2qa内力及内力图小结内力:轴力、扭矩、剪力、弯矩1.求解的力学依据:平衡理力：刚体的平衡，求约束反力(外力)材力：构件一部分的平衡，求内力(截面法)代数方程微段的平衡微分关系式微段的平衡：微分关系式弹性力学：微体的平衡平衡微分方程研究对象：刚化原理、刚体的平衡方程(从这个角度认识问题，就不必再一一归纳：在变形体力学中，力的可传性原理适用吗?力系是否可简化等问题)2.(从)数学(角度)：内力函数及其图象(1)内力符号（a）FN、T、FS与坐标无关，适用于刚架。（b）M与坐标相关(凹凸性暗含了坐标上指还是下指)，画在受压侧，物理属性与坐标无关。(2)作图先求内力函数(FS、M方程，T、FN方程)端值、极值、正负号刚架看作分段的梁()()直接作图也可由截面法段值积分关系连线微分关系Thanks！Thanks！