弯曲变形doulan9

1、Longitudinal deformation l;Twist deformation ;Bending deformation?vElasticalDeflection curve/Nl EAn/pM l GI,pzEA GIEIstiffness/zzM l EIdeflection curve function v = v ( x )rotating angle function = ( x )v: deflection or displacement : rotating angle or slopedtan( )dvv xxddvx Review1zME I2 3 2(1)vv)(

2、xv ZIExMxv)()( Basic differential equation of elastical curve ( )( )dZM xvxxCEI( )( )d dZM xv xx xEICxD( )( )dZEI vxM xxC( )( )d dZEI v xM xx xCxDBoundary conditionKnown deflectionand rotating angle例例6.1 Determine deflection curve function of beam shown in fig.22)(xqxMMoment function32( )d26ZqqxEI v

3、 xxxCC 43( )d624ZqqxEI v xxxCxDCxD 0,0:vvlx63qlC 84qlD)43(24)(434xxllEIqxvz)(6)(33xlEIqxvzxl qEIz)(84maxzEIqlv)(63maxmaxzEIqlvBoundary conditonlFbRAlFaRBxlFbxRxMA)(1)0(ax)()(2axFxRxMA)(axFxlFb)(lxa1212CxlFbvEIZ11316DxCxlFbvEIZRARB332)(66axFxlFbvEIZ22DxC2222)(22CaxFxlFbvEIZ021 DD)(62221bllFbCC0:,0:02

4、1vlxvx2121,:vvvvaxabFlBoundary conditoncontinuity conditon例例6.2)(62221blxxEIlFbvZ)()(632232axblblxxEIlbFvZaFv1v2ABCbIf a = b)(483maxZEIFlv2max16ZFlEIIf a b)(39)(322maxlEIblFbvZ3220blxIf b = 0Fb = M00.5773lxl)(392maxZEIMlvZEIMllvv16)221 （中maxmax2.64vvv中（2）ACCv,)()(mvFvvCCC)(48)(3ZCEIFlFv)()(mFCCC)(24

5、)(2ZCEIFlm)()(mFAAA)(16)(2ZAEIFlF)(48)(2ZAEIFlm)(24)481161(22ZZEIFlEIFlFABCm=Fl/2l/2l/2FABCl/2l/2ABCml/2l/2例例6.3 Found vC and AExam 6.4 Found vC of beam.Exchange loading )(8)2(41EIaqvCavvBBC222)(842EIqavB)(632EIqaB)(2474EIqa21CCCvvv21CCCvvv)(24412472444EIqaEIqaEIqaABCaaqABCqABCqBqACvB2 B2Rigidizatio

6、n in SequeceAssume BC is rigid, AB deformedavvBBC111)(127234231EIqaEIaMEIaQvBBB)(12194111EIqaavvBBC)(23221EIqaEIaMEIaQBBB Assume BC deformed , AB is rigid)(842EIqavC)(2441421EIqavvvCCCABCaaqBACqvC 1aaAa B1vB1QBMBvC2ACBqvC1)(6(6dd2xaxEIxqvC)(2441d42EIqavvaaCCEquivlently replace the loading )(4881)236

7、(6)23(42EIqaaaEIaqavC2 . 1828182Element finer, precision higherin x section dF = q dx , integrateABCaaqABCaaxqdxAC3a/2F=qaExam 6.5 found vC()/2CABvvvaaABqCABEI2EICaaaaABC2ql2qlvC2qlACABq/2C0,ACCAvvvACvvCABqC4qa34ql121CCCCvvvvExam 6.6 Found support reactionEquilibrium equationqlRRAC22Constrained cond

8、ition021CCCvvv)(245384)2(5441EIqlEIlqvCPhysical equation )(648)2(332EIlREIlRvCCCComplementaryequation0624534EIlREIqlCSolved)(45 qlRC)(83qlRRBAqABEICllRARBRCCqll1Cv2CvRCllComplementaryequationABCaaFABClaaEIEADqABClaaFDqlFC Exam 6.7 Found support reaction0DvCDClvAElFlCD)()(qvFvvCCCEIaFFvC3)2()(3EIqaqv

9、C2441)(3EIqaEAFlEIFavD244138430CDClv43412483qaIFalIAConstrained conditionPhysical equation lvmaxmax（1）Change loading and support（2）Reduce span length；（3） Optimize shape of section 11()1000700l（4） Optimize shape of beam zWxMx)()(max)(xMM )()(xMxWvariable section beamF 6)(2xFhxb2 6)(hxFxb6)(bxFxhstepp

10、ed shaftsuperposed beam轴力轴力N扭矩扭矩Mn弯矩弯矩M剪力剪力QNAnpMIzMyI NA npMW zMWNllEAEnpM lGIlGzMlEI1zMEIv/l v材料的材料的力学性能力学性能拉压实验拉压实验扭转实验扭转实验截面的截面的几何性质几何性质静矩与形心静矩与形心惯性矩与惯性矩与惯性积惯性积变形几何变形几何变形物理变形物理拉压变形拉压变形 l扭转变形扭转变形 弯曲变形弯曲变形挠度挠度v, y截面形心的竖向位移截面形心的竖向位移转角转角 截面绕中性轴转过的角度截面绕中性轴转过的角度v6-1 概概 述述挠度曲挠度曲线线挠度曲线方程挠度曲线方程 v ( x )转角

11、方程转角方程 ( x )dtan( )dvv xxddvx1zMEI2 3 21(1)vv)(xv ZIExMxv)()( 挠曲线挠曲线基本微分方程基本微分方程( )( )dZM xvxxCEI( )( )d dZM xv xx xEICxD( )( )dZEI vxM xxC( )( )d dZEI v xM xx xCxD边界条件边界条件v挠度曲线挠度曲线已知的挠度和转角已知的挠度和转角解：解：lFbRAlFaRBxlFbxRxMA)(1)0(ax)()(2axFxRxMA)(axFxlFb)(lxa1212CxlFbvEIZ11316DxCxlFbvEIZRARB332)(66axFxl

12、FbvEIZ22DxC2222)(22CaxFxlFbvEIZ021 DD)(62221bllFbCC0:,0:021vlxvx2121,:vvvvaxabFl边界条件边界条件连续条件连续条件ba极值点在极值点在 AC 段段)(39)(322maxlEIblFbvZba 极值点在极值点在 C 点点)(483maxZEIFlv0bFb 形成一个力偶形成一个力偶 M极值点极值点llx577.0303220blx)(392maxZEIMlv中点挠度中点挠度ZEIMllvv16)221 （中相对误差相对误差中64. 2maxmaxvvvaFv1v2ABCbaABbFM例例6-3 求梁指定截面求梁指定截

13、面 的挠度和转角。的挠度和转角。（1）Acv,解：解：)(245385)2(5441ZZCEIqaEIaqv)(648)2(432ZZCEIqaEIaFv)(83421ZCCCEIqavvv)(324)2(331ZZAEIqaEIaq)(216)2(332ZZAEIqaEIaqaaF = q aABqCqaaABCaaFABC)(65321ZAAAEIqa例例 6.4 求图示悬臂梁的求图示悬臂梁的 vC解一：增减载荷解一：增减载荷)(8)2(41EIaqvCavvBBC222)(842EIqavB)(632EIqaB)(2474EIqa21CCCvvv21CCCvvv)(24412472444

14、EIqaEIqaEIqaABCaaqABCqABCqBqACvB2 B2解二：逐段刚化法解二：逐段刚化法先假设先假设 BC 段刚性，只有段刚性，只有 AB 段变形段变形avvBBC111)(127234231EIqaEIaMEIaQvBBB)(12194111EIqaavvBBC)(23221EIqaEIaMEIaQBBB再考虑再考虑 BC 段的变形段的变形 ( AB 段刚性段刚性 )(842EIqavC)(2441421EIqavvvCCCABCaaqBACqvC 1aaAa B1vB1QBMBvC2ACBqvC1边界条件边界条件连续条件连续条件挠度挠度v, y转角转角 弯曲弯曲变形变形的度

15、量的度量曲率曲率1/ 1zMEIZIExMxv)()( vv直接直接积分法积分法( )( )dZEI vxM xxC( )( )d dZEI v xM xx xCxD查表查表叠加法叠加法直接叠加直接叠加增减载荷增减载荷逐段刚化逐段刚化载荷代换载荷代换结构代换结构代换例例6-5 求图示简支梁的求图示简支梁的 vC()/2CABvvv解：aaABqCABEI2EICaaaaABC2ql2qlvC2qlACABq/2C0,ACCAvvv解：ACvvCABqC4qa34ql121CCCCvvvv(B)(C)(A)(D)顶针顶针例例6-6 求下列超静定梁的支反力求下列超静定梁的支反力解：解除多余约束代之以约束反力解：解除多余约束代之以约束反力平衡方程平衡方程qlRRAC22约束条件约束条件021CCCvvv)(245384)2(5441EIqlEIlqvC物理方程物理方程)(648)2(332EIlREIlRvCCC补充方程补充方程0624534EIlREIqlC解得解得)(45 qlRC)(83qlRRBAqABEICllRARBRCCqll1Cv2CvRCll解：解除多余约束代之以约束反力解：解除多余约束代之以约束反力约束条件约束条件0DvCDClvABCaaFAElFlCD)()(qvFvvCCCEIaFFvC3)2()