1、Mechanics of Materials41 Concepts of planar bending and calculation sketch of the beam42 The shearing force and bending moment of the beam43 The shearing-force and bending-moment equations the shearing-force and bending-moment diagrams44 Relations among the shearing force、the bending moment and the
2、density of the distributed load and their applications 45 Plot the bending-moment diagram by the theorem of superpositiom46 The internal-force diagrams of the planar rigid theorem frames and curved rods Exercise lessons about the internal force of bending CHAPTER 4 INTERNAL FORCES IN BENDING 41 平面弯曲
3、的概念及梁的计算简图平面弯曲的概念及梁的计算简图42 梁的剪力和弯矩梁的剪力和弯矩43 剪力方程和弯矩方程剪力方程和弯矩方程 剪力图和弯矩图剪力图和弯矩图44 剪力、弯矩与分布荷载集度间的关系及应用剪力、弯矩与分布荷载集度间的关系及应用45 按叠加原理作弯矩图按叠加原理作弯矩图46 平面刚架和曲杆的内力图平面刚架和曲杆的内力图 弯曲内力习题课弯曲内力习题课第四章第四章 弯曲内力弯曲内力 41 CONCEPTS OF PLANAR BENDING AND CALCULATION SKETCH OF THE BEAM1、CONCEPTS OF BENDING1).BENDING:The actio
4、n of the external force or external the couple vector perpendicular to the axis of the rod makes the axis of the rod change into curve from original straight lines,this deformation is called bending.2).BEAM：The member of which the deformation is mainly bending is generally called beam.41 平面弯曲的概念及梁的计
5、算简图平面弯曲的概念及梁的计算简图一、弯曲的概念一、弯曲的概念1.弯曲弯曲:杆受垂直于轴线的外力或外力偶矩矢的作用时,轴 线变成了曲线，这种变形称为弯曲。2.梁：梁：以弯曲变形为主的 构件通常称为梁。3).3).Practical examples in engineering about bending3.3.工程实例工程实例4).4).Planar bending：After deformation the curved axis of the beam is still in the same plane with the external forces.Symmetric bendin
6、g（as shown in the following figure）a special example of the planar bending.The plane of symmetryMP1P2q4.4.平面弯曲：平面弯曲：杆发生弯曲变形后，轴线仍然和外力在同一 平面内。对称弯曲（如下图）平面弯曲的特例。纵向对称面纵向对称面MP1P2qUnsymmetrical bending if a beam does not possess any plane of symmetry,or the external forces do not act in a plane of symmetry
7、 of the beam with symmetric planes,this kind of bending is called unsymmetrical bending.In later chapters we will mainly discuss the bending stresses and deformations of the beam under symmetric bending.非对称弯曲 若梁不具有纵对称面，或者，梁虽具有纵 对称面但外力并不作用在对称面内，这种 弯曲则统称为非对称弯曲。下面几章中，将以对称弯曲为主，讨论梁的应力和变形计算。2、Calculation
8、sketch of the beam In general supports and external forces of the beam are very complex.We should do some necessary simplification for them for our convenient calculation and obtain the calculation sketch.1).Simplification of the beams In general case we take the place of the beam by its axis.2).Sim
9、plification of the loads The loads(including the reaction)acting on the beam may be reduced into three types：concentrated force、concentrated force couple and distributed force.3).Simplification of the supports二、梁的计算简图二、梁的计算简图 梁的支承条件与载荷情况一般都比较复杂，为了便于分析计算，应进行必要的简化，抽象出计算简图。1.构件本身的简化构件本身的简化 通常取梁的轴线来代替梁。
10、2.载荷简化载荷简化 作用于梁上的载荷（包括支座反力）可简化为三种类型：集中力、集中力偶和分布载荷。3.支座简化支座简化Fixed hinged support 2 constraints，1 degree of freedom.Such as the fixed hinged support under bridges，thrust ball bearing etc.Movable hinged support 1 constraint，2 degree of freedom.Such as the movable hinged support under the bridge，ball b
11、earing etc.固定铰支座 2个约束，1个自由度。如：桥梁下的固定支座，止推滚珠轴承等。可动铰支座 1个约束，2个自由度。如：桥梁下的辊轴支座，滚珠轴承等。Rigidly fixed end 3 constraints，0 degree of freedom.Such as the support of diving board at the swimming pool，support of the lower end of a wooden pole.XAYAMA4)Three basis types of beams Simple beam(or simply supported b
12、eam)M Concentrated Concentrated force coupleforce coupleq(x)Distributed forceDistributed forceCantilever beamA固定端 3个约束，0个自由度。如：游泳池的跳水板支座，木桩下端的支座等。XAYAMA4.梁的三种基本形式梁的三种基本形式简支梁M 集中力偶集中力偶q(x)分布力分布力悬臂梁Overhanging beamConcentratedConcentratedforceforcePqUniformly Uniformly distributed distributed forcefor
13、ce5).Statically determinate and statically indeterminate beamsStatically determinate beams：Reactions of the beam can be determined only by static equilibrium equations，such as the above three kinds of basic beams.Statically indeterminate beams：Reactions of the beam cannot be determined or only part
14、of reactions can be determined by static equilibrium equations.外伸梁 集中力集中力Pq 均布力均布力5.静定梁与超静定梁静定梁与超静定梁静定梁：由静力学方程可求出支反力，如上述三种基本 形式的静定梁。超静定梁：由静力学方程不可求出支反力或不能求出全 部支反力。Example 1 1 A stock tank is shown in the figure.Its length is L=5m，its inside diameter is D=1m，thickness of its wall is t=10mm.Density of
15、steel is 7.8g/cm.Density of the liquid is 1g/cm.Height of the liquid is 0.8m.Length of overhanging end is 1m.Try to determine the calculation sketch of the stock tank.Solution：qUniformlyDistributed force 例例11贮液罐如图示，罐长L=5m，内径 D=1m，壁厚t=10mm，钢的密度为：7.8g/cm，液体的密度为：1g/cm,液面高 0.8m，外伸端长 1m，试求贮液罐的计算简图。解：解：q
16、均布力均布力LgLAgLALgVLmgq2211rad855131060.gRRgDt2221)sin(21gAgA2211(kN/m)99.81000)sin106.3(1.8550.521 0.53.14897800010114322.qUniformlyDistributed forceLgLAgLALgVLmgq2211rad855131060.gRRgDt2221)sin(21gAgA2211(kN/m)99.81000)sin106.3(1.8550.521 0.53.14897800010114322.q 均布力均布力42 THE SHEARING FORCE AND BENDI
17、NG MOMENT OF THE BEAM1、Internal force in bending：Example Knowing conditions are P，a，l,as shown in the figure.Determine the internal forces on the section at the distance x to the end A.PaPlYAXARBAABBSolution：Determine external forceslalPYYlPaRmXXABAA)(,0 ,00 ,042 梁的剪力和弯矩梁的剪力和弯矩 一、弯曲内力：一、弯曲内力：举例举例已知：
18、如图，P，a，l。求：距A端x处截面上内力。PaPlYAXARBAABB解：求外力lalPYYlPaRmXXABAA)(,0 ,00 ,0ABPYAXARBmmxDetermine internal forces method of sectionxYMmlalPYQYACA ,0)(,0AYAQMRBPMQInternal forces of the beam in bendingShearing forceBending moment1).Bending moment：M Moment of the internal force couple with the acting plane i
19、n the cross-section perpendicular to the section when the beam is bending.CCABPYAXARBmmx求内力截面法xYMmlalPYQYACA ,0)(,0AYAQMRBPMQ 弯曲构件内力剪力弯矩1.弯矩：M 构件受弯时，横截面上其作用面垂直于截面的内力偶矩。CC2).Shearing force：Q Internal force which the acting line in the cross-section parallel to the section,when the beam is bending.3).
20、Sign conventions for the internal forces:Shearing force Q:It is positive when it results in a clockwise rotation with respect to the object under consideration,otherwise it is negative.Bending moment M：It is positive when it tends to bend the portion concave upwards,otherwise it is negative.Q(+)Q()Q
21、()Q(+)M(+)M(+)M()M()2.剪力：Q 构件受弯时，横截面上其作用线平行于截面的内力。3.内力的正负规定:剪力Q:绕研究对象顺时针转为正剪力；反之为负。弯矩M：使梁变成凹形的为正弯矩；使梁变成凸形的为负弯矩。Q(+)Q()Q()Q(+)M(+)M(+)M()M()Example 2：Determine the internal forces acting on sections 11 and 22 section as shown in fig.(a).qLQQqLY11 0Solution：Determine internal forces by the method of s
22、ection.Free body diagram of the left portion of section 11 is shown in fig.（b）.Fig.（a）1111 0)(qLxMMqLxFmiA2、ExamplesqqLab1122qLQ1AM1Fig.（b）x1 例例2：求图（a）所示梁1-1、2-2截面处的内力。xyqLQQqLY11 0解：解：截面法求内力。1-1截面处截取的分离体 如图（b）示。图（a）1111 0)(qLxMMqLxFmiA二、例题二、例题qqLab1122qLQ1AM1图（b）x1L)axq Q22(axqMqLxFmiB0)(21,0)(2222
23、Free body diagram of the left portion of section 22 is shown in fig.（b）.)ax(qQqLY0222222)(21qLxaxqMxy图（a）qqLab1122qLQ2BM2x2图（c）L)axq Q22(axqMqLxFmiB0)(21,0)(22222-2截面处截取的分离体如图（c）)ax(qQqLY0222222)(21qLxaxqMxy图（a）qqLab1122qLQ2BM2x2图（c）1.Internal-force equations:Expressions that show the internal force
24、s as functions of the position x of the section.2.The shearing-force and bending-moment diagrams:)(xQQ Shearing force equation)(xMM Bending moment equation)(xQQ Shearing-force diagramsketch of the shearing-force equation)(xMM Bending Moment diagramsketch of the bending-moment equation43 THE SHEARING
25、-FORCE AND BENDING-MOMENT EQUATIONS THE SHEARING-FORCE AND BENDING-MOMENT DIAGRAMS1.内力方程：内力与截面位置坐标（x）间的函数关系式。2.剪力图和弯矩图：)(xQQ 剪力方程)(xMM 弯矩方程)(xQQ 剪力图的图线表示)(xMM 弯矩图的图线表示43 剪力方程和弯矩方程剪力方程和弯矩方程 剪力图和弯矩图剪力图和弯矩图Example 3 3 Determine the internal-force equations and plot the diagrams of the beam shown in the
26、 following figure.PY)x(QOSolution：Determine the reactions of the supports)Lx(PMxY)x(MOO Write out the internal-force equationsPL MPYOO;PPlot the internal-force diagrams Q(x)M(x)xxPPLYOLM(x)xQ(x)MO 例例3 3 求下列各图示梁的内力方程并画出内力图。PY)x(QO解：求支反力)Lx(PMxY)x(MOO 写出内力方程PL MPYOO;PYOL根据方程画内力图M(x)xQ(x)Q(x)M(x)xxPPLM
27、OSolution：Write out the internal-force equationsPlot the internal-force diagramqx)x(Q221qx)x(MLqM(x)xQ(x)Q(x)x qL22qLM(x)x解：写出内力方程根据方程画内力图qx)x(Q221qx)x(MLqM(x)xQ(x)Q(x)xM(x)x qL22qL)3(6220 xLLq)x(QSolution：Determine the reactions of the supports Write out the internal-force equations3;600Lq RLqRBAq0
28、RA Plot the internal-force diagrams RBL)xL(LxqxM2206)(xL33Q(x)x620Lq30Lq27320LqM(x)3(6220 xLLq)x(Q解：求支反力内力方程3;600Lq RLqRBAq0RA根据方程画内力图RBL)xL(LxqxM2206)(xL33Q(x)x620Lq320Lq27320LqM(x)1 1、Relations among the shearing force、the bending moment and the the distributed loadBy analysis of the equilibrium o
29、f the infinitesimal length dx，we can get0dd0)x(Q)x(Qx)x(q)x(QY)x(Qx)x(qdd 44 RELATIONS AMANG THE SHEARING FORCE,THE BENDING MOMENT AND THE INDENSITY OF THE DISTRIBUTED LOAD AND THEIR APPLICATIONS dxxq(x)q(x)M(x)+d M(x)Q(x)+d Q(x)Q(x)M(x)dxAy xqxxQddSlope of the tangential line at a point in the shea
30、ring-force diagram is equal to the intensity of the distributed load at the same point.一、一、剪力、弯矩与分布荷载间的关系剪力、弯矩与分布荷载间的关系对dx 段进行平衡分析，有：0dd0)x(Q)x(Qx)x(q)x(QY)x(Qx)x(qdd 44 剪力、弯矩与分布荷载集度间的关系及应用剪力、弯矩与分布荷载集度间的关系及应用dxxq(x)q(x)M(x)+d M(x)Q(x)+d Q(x)Q(x)M(x)dxAy xqxxQdd剪力图上某点处的切线斜率剪力图上某点处的切线斜率等于该点处荷载集度的大小。等于
31、该点处荷载集度的大小。q(x)M(x)+d M(x)Q(x)+d Q(x)Q(x)M(x)dxAy0)(d)()()(d(21)d(0,)(2xMxMxMxxqxxQFmiA)(d)(dxQxxMSlope of the tangential line at a point in the bending-moment diagram is equal to the magnitude of the shearing force at the same point.)(d)(d22xqxxMRelation between the bending moment and the indensity
32、 of the distributed load：q(x)M(x)+d M(x)Q(x)+d Q(x)Q(x)M(x)dxAy0)(d)()()(d(21)d(,0)(2xMxMxMxxqxxQFmiA)(d)(dxQxxM弯矩图上某点处的切线斜率等于该点处剪力的大小。弯矩图上某点处的切线斜率等于该点处剪力的大小。)(d)(d22xqxxM弯矩与荷载集度的关系是：弯矩与荷载集度的关系是：2、Relations between the shearing force、the bending moment and the external loadExternal forceNo external-
33、force segmentUniform-load segmentConcentrated forceConcentrated coupleq=0q0q0QQ0q0QQ0q0QQ0q0QQ0 x斜直线增函数xQxQ降函数xQCQ1Q2Q1Q2=PxQC自左向右突变无变化斜直线xM增函数xM降函数xMxMxMxM曲线坟状盆状自左向右折角折向与P反向M1 M2自左向右突变与m反mMM21Example 1 Plot the bending-moment diagrams of the beam shown in the following figure.2PaaP=2PP+xMxM1xM2=+2P
34、a2PaPa(1)例例1 绘制下列图示梁的弯矩图。2PaaP=2PP+xMxM1xM2=+2Pa2PaPa(1)(2)aaqqqq=+xM1=xM+xM23qa2/2qa2/2qa2(2)aaqqqq=+xM1=xM+xM23qa2/2qa2/2qa2(3)PL/2L/2PL/2=+PxM2xM=+PL/2PL/4PL/2xM1+PL/2(3)PL/2L/2PL/2=+PxM2xM=+PL/2PL/4PL/2xM1+PL/2(4)50kN2m2m20kNm=+xM2xM=+20kNm50kNmxM120kNm50kN20kNm20kNm+20kNm30kNm20kNm(4)50kNaa20kN
35、m=+xM2xM=+20kNm50kNmxM120kNm50kN20kNm20kNm+20kNm30kNm20kNmyzhb)4(222yhIQbIQSzzzSolution:（1）Shearing stress on the cross section isExample 2 2 The structure is shown in the figure.Try to prove：(1）resultant of the shearing stresses in an arbitrary cross section is equal to the shearing force in the sa
36、me section；（2）Resultant moment of the normal stresses in an arbitrary cross section is equal to the bending moment in the same section；（3）which force can balance the resultant of the shearing stress in the longitudinal section at middle height balanced?.qNormal stress on the cross section is zIMyyzh
37、b)4(222yhIQbIQSzzz解：（1）横截面的剪应力为：例例22结构如图，试证明：（1）任意横截面上的剪应力的合力等于该面的剪力；（2）任意横截面上的正应力的合力矩等于该面的弯矩；（3）过高度中点做纵截面，那么，此纵截面上的剪应力的 合力由哪个力来平衡？q横截面的正应力为：zIMyhhzAybyhIQA5.05.022d )4(2dMIIMAIMyMzzh.h.zz50502d(2)Resultant shearing force in the cross section is：QhhIQbz)2(324233(3)Resultant force couple hhzAybyhIQA5
38、.05.022d )4(2dMIIMAIMyMzzh.h.zz50502d(2)横截面上的合剪力为：QhhIQbz)2(324233(3)合力偶)(bhqx.AxQ.51)(51maxhqLxqxhAQLLAB43d)(23d200zAWAMAN2211max1max 11AABNQ(4)Shearing stress in the middle longitudinal section is：Resultant of the shearing stress in the longitudinal section is balanced by resultant of the normal s
39、tress in the right-side section.(5)Resultant of the shearing stress in the longitudinal section is：max hqLbhbhqL4326221222xL)(bhqx.AxQ.51)(51maxhqLx)qx(hAQLLAB43d23d200zAWAMAN2211max1max 11AABNQ(4)中面上的剪应力为：纵面上的合剪力与右侧面的正应力的合力平衡。(5)纵截面上的合剪力大小为：max hqLbhbhqL4326221222xL Chapter 4 Exercises 1.Try to lis
40、t some members in bending 1.Try to list some members in bending according to your experiences and simplify them according to your experiences and simplify them into various kinds of beams.into various kinds of beams.2.Try to plot the bending-moment diagram of 2.Try to plot the bending-moment diagram
41、 of the beam by the superposition method.the beam by the superposition method.222222 第四章第四章 练习题练习题 一、试根据自己的实践经验，列举一些弯曲变一、试根据自己的实践经验，列举一些弯曲变形的构件，并将它们简化为各种类型的梁。形的构件，并将它们简化为各种类型的梁。二、试用叠加法作梁的弯矩图。二、试用叠加法作梁的弯矩图。22222223.Plot the Q-diagram and the M-diagram of the 3.Plot the Q-diagram and the M-diagram of
42、the composite beam with a middle hinge shown in the composite beam with a middle hinge shown in the figure.figure.qa2qa2/2 三、作图示具有中间铰的组合梁的三、作图示具有中间铰的组合梁的Q Q、M M图。图。qa2qa2/2 4.Lift an equal-section beam with the weight q(N/m).Ask:what is the reasonable location x of the lift point?Note:make the maxim
43、um positive bending moment equal to the absolute value of the maximum negative bending moment.Solution:As We have Take (It is meaningless for x to take the negative value.)22qxMA)2(242xLqLLLqMCCAMM04422LLxxLx221Lx207.0ql/2ql/2 四、起吊一根自重为四、起吊一根自重为q(N/m)的等截面梁。问：的等截面梁。问：起吊点位置起吊点位置x应为多少才最合理？提示：使梁的最应为多少才最合理？提示：使梁的最大正弯矩与最大负弯矩的绝对值相等。大正弯矩与最大负弯矩的绝对值相等。解：解：取取 （x为负值无意义）为负值无意义）22qxMA)2(242xLqLLLqMCCAMM04422LLxxLx221Lx207.0ql/2ql/2
