第13章 能量法 ppt课件.ppt

1、)d(d11lFW EAlFl11d)d( FF lF lFO l l1d l1dF1F1lFEAlFFEAlFWWF22dd2101 lFlFWV2121N EAlFEAFllN EAlFEAlFV222N2 niiiiiAElFV12N2 lxEAxxFV02N)(2)d(AllFVU2121 E 222122EE 或或p2p2epeee222121GIlTGIlMGIlMMMWV lxxGIxTVd)(2)(p2 niiiiiIGlTV1p22 EIlMEIlMMMWV221212 eeexxEIxMVld)(2)(2e xxEIxMxxGIxTxxEAxFVllld)(2)(d)(2)

2、(d)(2)(2p22N dxdydzxyzabd )dd(d21)d)(dd(21dzyxxzyW zyxWVWVV21dddddddd dxdydzxyzabd 21 GG2222 l AVxAVVddd GG2222 p222p2p22d)(2 dd2)(dd2GIlTAITGlxAGITxAGVAl Al A p2e2GIlMV pepGIlMGITl 2eMV123FV21 F3BCF2AF1)(233221123322112FFCCFFCFFCFCFC F3ABCF1F2123 iF3ABCF1F2123)(21332211FFFV ixFxM )(EIlFxEIFxxEIxMVl

3、l6d2)(d2)(32022BwFW 21EIFlwB33 ACEIlbaFbEIlaFaEIlbFxEIxlFaxEIxlFbxEIxMVbal63232d2)(d2)(d2)(22232223222202210212 CwFW 21EIlbFawC322 EIRFREIFRREIMVl8d2)sin(d2)(322022 sin)(FRM yFW 21EIFRy43 ABCabF1F2ABCabF1F211BF aEA EAaFFWB22121111 ()22222122CFabWFEA ()22CF abEA ABCabF1F222BF aEA 12312BF F aWFEA ()11

4、2212221212112222BCBVWFFFF aFabF F aEAEAEAABCabF1F2EAbaF2)(22 EAaF221ABCabF1F2EAaF1EAaFF21()112212221212112222BCBVWFFFF aFabF F aEAEAEAMe23e14816M lFlEIEI 2e2163M lFlEIEI ACBFl/2l/2()222 3ee1e21112296616M lM FlF lVFMEI3148FlEI 31112248FlWFFEI EIlM3e EIlMMMW32121eee2 ACBFl/2l/2ACBFl/2l/2Me2e316M lEI 2

5、e3316M lWFFEI ACBl/2l/2EIlMMEIlMFEIFlFV321164821ee2e3 ACBFl/2l/2Me 1 2F1F2F1F2 1 2F3 3 4F411221122FF33441122FFF1F2 1 2F3 3 4F42 1 FF2211 FFFFFFV221144332211121212121 F1F2 1 2 3 4F4F3FFFFFFV443344332211221212121 21VV FFFF44332211 4 3 FF3311 31 2 1 3F1F2F3 332211212121FFFWV,.12312iiF.1122iiFFF.112212i

6、iiiVFFFF 2 1 3F1F2F312iiF.1122iiVFFF.1122iiiiFFFF iiVF iiVFiiFV iiFV xFxFEAxFEAxxFFFViiiid)()(2)d(NN2N xFxTGIxTGIxxTFFViiiid)()(2)d(pp2 xFxMEIxMEIxxMFFViiiid)()(2)d(2 ijnjjjiiFFEAlFFV N1NiiFV 2)d(2)d(2)d(2p22N llliEIxxMGIxxTEAxxFFxFxMEIxMxFxTGIxTxFxFEAxFiiid)()(d)()(d)()(pNN FABCMelae1e11)()(MxlFalM

7、xM 111)(xlaFxM 1)(1e11 lxMxM222)(FxxM 0)(e22 MxM222)(xFxM ABClaFx1x2eRAMFaFllMe 1011d)()(xFxMEIxMwlCABClaFx1x220e222210e11d)()(d)()(xMxMEIxMxMxMEIxMalA 202222d)()(xFxMEIxMa )( )363(13e2 FalaMFlaEI)63(1eFlalMEI Me解：解： 1 1. . 在在x1 处施加集中力处施加集中力F，列弯矩方程列弯矩方程xFLxLxqxqlMMxMFq12122)(求图示梁的挠曲线方程。求图示梁的挠曲线方程。11

8、1( )()MxlxxxxFl11( )MxLxxFLxABqlx1CAC段：段：)(22)(112xxFxFlxlxqxqlMMxMFqCB段：段：F3.3.求挠曲线方程求挠曲线方程2.2.求求应变能应变能xEIxMxEIxMxEIxMVlxx ld2)(d2)(d2)(1212021 211 11221 0 00( )()ddx lxFFVMMMxMw xxxFEIFEIF01111201121d)()(22d22)(11FlxxxxxxlxlEIxxFxFlxlxqxqlxxlxlEIxFlxlxqxqlxwxxxxlxlEIxqxqlxxlxlEIxqxqlxwlxxd)(22d22)

9、( 1120 12111)2(24)(322xlxlqxxwABCDaa2aMe)(21 eaRRMMaFFFAyD FRDFRAxFRAyFFAx RxMMaFxM)(21)(ae xFxM )(axMxM2)(a aMMaFxM2)(21)(ae xaFxM 2)(0)(a MxMxFxM )(0)(a MxMxxABCDaa2aMexFxM aFxxM )(FRDFRAxFRAy2axxABCDaaMe00a FMxFV00aa FMCMV axxEI0d01 axxaMEI0e)d2(1 axxaxMEI20ed21)(6172e EIaM0d00dd2212000ee aaaxxMx

10、axaxMEIEIaM32e FRDFRAxFRAyBCllqxFxM )(2)(2qxFxxM BlQMBxBCllqFxxqlFF 22qlFlMB xqlFQxxM)()( xFxM )(2)(2qlFlMxTB lFxT )(0 FiFV644dI 324pdI llxFxTGIxTxFxMEIxMxFxMEIxM0p0d)()(d)()(d)()( lllxlqlGIxxqlxEIxxqxEI02p002d21d1)d)(2(1)(2411p44 GIqlEIqlMaxF1FABCDll2 la)(MFxxM xFxM )(1)(a MxMF1ABCF2FlF1xFABCDll2 l

11、Max1)(a MxM1)(a MxMlFxM2)( lFxM2)( a2)(MFlxM xFMFlxM1a2)( F1FABCDll2 lxF2FlxEIFl2132 d1)2(d121001 llxxFFlxFlEIFFMDMV 1a0a lxFxEI20d11MaMa 例例13-1013-10计算图（计算图（a）所）所示开口圆环在示开口圆环在 P力作用下力作用下切口的张开量切口的张开量 AB ，EI=常常数。数。)cos1 ( FRMdF FF FF FF FFNMFQcosFFNFFNcos)cos1(RFMxFxMEIxMFViiid)()(022d)22(2REAFIEMVNd)c

12、os)cos1 (22022REAFIEFRABEAFREIFRAB330d)(2RFFEAFFMIEMFVNNAB2434/3ErFRErFRAB)121 (4/32243RrErFRAB0008. 012, 1 . 022RrRr注意注意:（2 2）所求截面位移无相应的载荷时要施加该载荷，按）所求截面位移无相应的载荷时要施加该载荷，按卡氏第二定理求导后令假设的载荷为零。卡氏第二定理求导后令假设的载荷为零。（1 1）F F视为广义力，即在不同的载荷下它分别代表集中视为广义力，即在不同的载荷下它分别代表集中力、力对、力矩等。相应的力、力对、力矩等。相应的 视为广义位移，在不同的视为广义位移，在

13、不同的载荷下它分别代表集中力方向的线位移、力对作用点的载荷下它分别代表集中力方向的线位移、力对作用点的相对位移、力矩转向上的转角等。相对位移、力矩转向上的转角等。（3）如果结构上作用几个相同的载荷，则应分别给出）如果结构上作用几个相同的载荷，则应分别给出不同的标识，不同的标识，按卡氏第二定理求导后令它们取原值。按卡氏第二定理求导后令它们取原值。（4）应变能积分中的内力函数式不可展开，且先求导后）应变能积分中的内力函数式不可展开，且先求导后再代如积分号内运算。再代如积分号内运算。 例例13-1113-11 由于约束反力处的位移是已知的，所以卡氏由于约束反力处的位移是已知的，所以卡氏第二定理建立在

14、该约束处的位移协调方程，从而第二定理建立在该约束处的位移协调方程，从而求解超静定问题。求解超静定问题。 力法：以未知力作为基本未知函数，利用有关力法：以未知力作为基本未知函数，利用有关的定理及变形关系求解超静定问题称为力法。的定理及变形关系求解超静定问题称为力法。A132lF F解解：（：（1 1）平衡方程）平衡方程取节点取节点A分析（如图）分析（如图）F1NF2NF3NFA Asinsin:031NNxFFF213coscos:0NNNyFFFFFA132F FFN213sinsinNNFF21)cos(sinNNFFctgF)(sin21ctgctgFPFNN)(sin23ctgctgFP

15、FNNEAlFEAlFEAlFVNNN222323222121121NFN123NNFF（3 3）求约束反力）求约束反力0coscos2332211NNNNNNNBFFEAlFEAlFFFEAlF0coscos321NNNFFF0)(sincos)(sincos22222ctgctgFFFctgctgFFNNN1cossincossin)(coscossinsin222222ctgctgFFN：若1cos232NFF1cos232FFN0ie WW 1 2 3 4AlBF4F1F2F3FRAFRBiiiBAiiiFRRFW 4141e00dxMM+dMd2 d2 F4F1F2F3FRAAlFR

16、BBdxFSMFS+dFSM+dM2d)d(2d MMMdxdxd2 d2 d2 d2 2d)d(2dSSSFFF F4F1F2F3FRAAlFRBBdxdxFSMFS+dFSM+dMFMddS 2d)d(2d MMM2d)d(2dSSSFFF 0)dd(dddSiei FMWWW)dd(dSiFMW lSlii)dd(dFMWW0)dd(lS41 FMFiii lNS41)dddd( TFFMFiii A lxEIxMVd2)(2 lxEIxMVd2)(2AwVVV 11AF1F2=1F0AF1F2wA)()(xMxM lxEIxMxMVd2)()(2212VV lAxEIxMxMwVVd2

17、)()(12 llllAxEIxMxMxEIxMxEIxMxEIxMxMwVVd)()(d2)(d2)( d2)()(1222 lAxEIxMxMwd)()( lxEIxMxMd)()( lxEIxMxMd)()( niiiiEAlFF1NN lllxEIxMxMxGIxTxTxEAxFxFd)()(d)()(d)()(pNNAqBCll/2ql/2ql/222)(2qxxqlxM )(0lx AB11/21/2CxxxM21)( )2(0lx )(3845)d22(22)d()(4220 EIqlxqxxqlEIxEIxxMxMwl/lCAqBCll/2ql/2ql/211)( xlxM)(

18、0lx EIqlxqxxqllxEIEIxxMxMllA24)d221)(1)d()(320 AqBCll/2ql/2ql/2BACqF=qaa2aBAABCa2a1xCqF=qaa2aFRAx1/22RqaFA 22)(2qxxqaxM 2)(xxM xqaxM )(xxM )()(32 )d)()d2)(22(140202 EIqaxxqaxxxqxxqaEIwaaCBAABCa2aCqF=qaa2aFRA1/2xx1BA1xxABCa2axCqF=qaa2ax1/2a22)(2qxxqaxM axxM2)( xqaxM )(1)( xMEIqaxqaxxaxqxxqaEIaaC65 )(

19、1)d()d2)(22(130202 FRAaABCFlEI1EI2BClEI1EI2a1aABCFlEI1EI2xxFxxM )(FaxM )(xxM )(axM )(ABC1lEI1EI2xxa)(3 )d( )(1 )d( )(1 d)()(d)()(221302010201 EIlFaEIFaxaFaEIxxFxEIxEIxMxMxEIxMxMlalay0)( xM1)( xMFxxM )(FaxM )(ABCFlEI1EI2xxaABClEI1EI2xxa202010201(1)d)(1 (0)d)(1d)()(d)()(EIFalxFaEIxFxEIxEIxMxMxEIxMxMla

20、laB 1CFabAB1abBACFab1abxxABBACCFxxM )(xxM )(0)(N xF0)(N xFxxFaxM )(axM )(FxF )(N1)(N xF llCxxFxFEAxxMxMEIH)d()(1)d()(1NN)(3d1)(1230 EAFbEIbFaEIFaxFEAb xaFaEIxxFxEIbad )(1d )(100Fab1abxxABBACCxxABCFabxxFxxM )(xxM )(0)( xT0)( xTABCFabABCabxxFxxM )(xxM )(FbxT )(bxT )(xxABCFabABCabxx( )( )( ) ( )llp11dd

21、CM x M xxT x T xxEIGI xxFxEIxxFxEIbad )(1d )(100 xbFbGIad )(10p p233)(3GIFabbaEIF )( A1A2BCllFFxlxM )(FlxM )(xxFxxM )(xxM )(FxxM )(xxM )(A1A2BCllFFxxxA1A2BCll11) (35)d)()d)(21300 EIFlx-l-Flx-x-FxEIllABCllqxxxx2)(2qxxM 2)(2qlxM xxM )(lxM )(ABCllqABCllq)(85)d2d2(140022 EIqlxlqlxxqxEIlly2)(2qxxM 2)(2ql

22、xM 0)( xMxxM )()(4)d2d02(140022 EIqlxxqlxqxEIllxxxxxABCllqABCllqxx2)(2qxxM 1)( xM2)(2qlxM 1)( xMEIqlxqlxqxEIllA32)d1)2(d1)2(130022 xxABCllqABCllq1FaaFABCDE132456789aFaaABCDE132456789aFaaFABCDE132456789a11 niiiiEAlFF1NNiFNiFNiliiilFFNNF2F221 / 21 / 21 / 21 / a2a22Fa/2Fa/2Fa/EAFa.EAFa)(EAlFFiiiiAC1242

23、3291NN BAORFF(a)BARPF d (b)BARP1(c)EIFRREIFRREIMMAB302203d)cos(12d)()(2 )cos(1)( FRM)cos(1)( RMsdOOxxMxMEIl)d()(1 xxMxMl)d()( ldxxCxCMMxxMxMl)d()( EIMxEIxMxMCl d)()( )ltandx M xx tanCCxM xCCM(x)xx)(xMlBxAxM )(xxMxBxxMAxxM)BxA(xxMxMllll)d()d()d(d )()(000 lxxM0)d(xxMl 0)d(M(x)xlx)(xMxCC lxxxM0)d(Clxxx

24、M 0)d(CcClllMxBAxBAxxMxBxxMAxxMxM )()d()d(d )()(00 CxBA M(x)xlx)(xMxcC d )()(1EIMxxMxMEIcl balhCClh顶点顶点3bl 3al 2hl 85l83llh 32lh顶点顶点clh顶点顶点c3hl 1 nhllnn2)1( 2 nlCqFC4lC11C221CM2CM82ql825l/ 2428323221qllql llMC 325485CqCF)(38453252424321 EIqllqlEIEIMEIMwCCC4lC11C221CM2CM82ql825l/ FCABalqFCABaalqMql2/

25、8Fa0)212322 322(132 aql-aFaaFalEIwC)(83alaqlF 1ABalCM ABFa/2a/2EIFaEIFaAB32)212123181(233 0 AB BAA)(838531324144 EIqa)(EIqaAHC5A300B500=1F01020 C5A300B500=1020 x1x1xxxEIxMxMxGIxTxTLLBd)()(d)()(1p11 FxxMAB )(xxMAB )(F.xTCA30)(1 30)(1.xTCA xEIFxxGI.F.dd303030021500p 343331020210403250306030101052103123060 .22mm8. p33GIlFllEIFlACABABAB