西南交大-材料力学-龚晖-弯曲内力

第四章弯曲应力§4-1对称弯曲的概念及梁的计算简图Ⅰ、弯曲的概念受力特点：MeMeABF几何特点：等直杆杆件受到垂直于杆轴线的外力（横向力）或外力偶（其矢量垂直于杆轴）作用。——以弯曲为主要变形的杆件通称为梁。梁2、任意两横截面绕垂直于杆轴的轴作相对转动。变形特点：1、直杆的轴线在变形后变为曲线；q3qa2a3aABC(1)求反力FAFB得：由0BMaFA323qa21qaFA35得：由0Y02BAFaqFqaFB31§4-2梁的剪力和弯矩•剪力图和弯矩图2(2)aq0简支梁Ⅰ、梁的剪力和弯矩q3qa2a3aABC(2)求截面内力11取1-1截面xFAFs1M1x00s1FFYA得：由qaFFA35s1001MxFMAC得：由qaxxFMA351FBFs1’3qa2M1’qaF35'1SqaxM35'1内力与截面位置有关剪力和弯矩的符号规则：q3qa2a3aABC22取2-2截面FBqFs2M2x3a-x得：由0Y2SFBFxaqF)3(2Sqaxaq31)3(得：由0CM2M)3()3(2122xaFxaqMB)3(31)3(212xaqaxaqBRq)3(xa021q2)3(xa)3(xaFB0q3qa2a3aABCqaFA35qaFB31qaFFA351sqaxxFMA351qaxaqF31)3(2s)3(31)3(2122xaqaxaqMxQoqa35qa31+xMo235qa234qa,031)3('2qaxaqM令ax38182maxqaM182qa+)(d)(dSxqxxF)(d)(dSxFxxM)(d)(d22xqxxM2112d)(SSxxxxxxqFF2112d)(SxxxxxxFMMⅢ、弯矩、剪力与分布荷载集度之间的关系及其应用FxMeq(x)分布荷载集度q(x)向上为正，向下为负FS0FS0FS=0突变折角无影响突变与x轴斜交极值MFSMqFS0M0FS0M=0FS0M0FS=0M=0F解：1）求内力例5-9求图示外伸梁的内力图。10kN/m5kNm2m5mABC1m5kN10kN/mDFAFB得：由0BM5AF523102111021150kN10AFkN15BF2）作剪力图xoFS(kN)1020qBCCBlSFFSS31010kN205qBDBrDlSFFSS1105kN55+-xoM(kNm)0SClFACMMS10kN/m5kNm2m5mABC1m5kN10kN/mDxoFS(kN)102055+-3）作弯矩图2100kNm2020集中M引起突变：逆时针向上突变顺时针向下突变15BCFCrBSMMS1m2m220211102115=0BDFBDSMMS0BM011maxSFCrSMM1102115kNm202022maxSFBSMM5.05210kNm25.11.25+解：kN81AyFkN29ByF例：试绘出图示有中间铰的静定梁的剪力弯矩图。1m0.5m1m3m1mBACDKEq=20kN/mMe=5kN·mF=50kNMAFAxFAyFBy1)求反力取CDKB为分离体，得：由0CM0555.2320ByF032050ByAyFF得：由0Y取整体为分离体，mkN5.96AM1m0.5m1m3m1mBACDKEq=20kN/mMe=5kN·mF=50kNMAFAxFAyFBy取AEC为分离体，得：由0CM05.0505.1AyAFM1m0.5mAEF=50kNMAFAxFAyFCyCFCx得：由0Y050CyAyFFkN31CyF1m0.5m1m3m1mBACDKEq=20kN/mMe=5kN·mF=50kNMAFAxFAyFBykN81AyFmkN5.96AMkN29ByF813129Fs图(kN)1.45m1m0.5m1m3m1mBACDKEq=20kN/mMe=5kN·mF=50kNMAFAxFAyFBy813129Fs图(kN)1.45m96.515.53155345M图(kN·m)§4-3平面刚架和曲杆的内力图解：0NxFCB段：(外侧受拉)qxxFS)20(2)(2axqxxM)20(axx例试作图示刚架的内力图。qF=2qaBCA2aa2aDFS(x)M(x)CqxFN(x)qF=2qaBCA2aa2aDy)0(2NayqayF(外侧受拉)BD段：0SyFayqayM022FS(x)M(x)qBCy2aFN(x)yF=2qaBCa2aDqFS(x)M(x))3(2NayaqayFDA段：)3(2SaxaqayFayaqayayqaqayM32)(222qF=2qaBCA2aa2aDFN(x)F=2qaBCA2aa2aDq2qaFN图0NxF)0(2NayqayF)3(2NayaqayF轴力符号的规定仍然有效。轴力图可画在杆件的任一侧，但应注明符。F=2qaBCA2aa2aDqqxxFS0SyF)3(2SaxaqayF2qa2qaFS图剪力符号的规定仍然有效。剪力图可画在杆件的任一侧，但应注明符。F=2qaBCA2aa2aDq)20(2)(2axqxxMayqayM022qayyM22qa26qa2M图弯矩不规定符号。弯矩图画在杆件的受拉侧。F=2qaBCA2aa2aDq2qaFN图2qa2qaFS图2qa26qa2M图解：π/20sinNFFmm例一端固定的四分之一圆环，半径为R，在自由端B受轴线平面内的集中荷载F作用如图，试作出其内力图。AOBRFFOhFN(FS(M(得：由0NFsinF0AOBRFFOhFN(FS(M(0h由得：SFcosF0π/20cosSFF得：由0CMMsinFR0π/20sinFRMＤAOBRFsinNFFBAFFN图cosSFFAFBFS图sinFRMFRABM图