第二章平面力系

1、第二章第二章 平面力系平面力系A(chǔ)F2F4F3F1FRF2F3F4FRFR1FR2F4F3F1FRF1AF2A F2F3F4FRF1A01niiF FF2F3FRF1AF4 PABCD ， FB FA P FBFAPPFB5 . 0tan PFPFBA2522 lhKMMN 2/sin sin2221PhPlFF Fh, 0 xABFOijy sincoscosFFFFFyx jiFFFyxyxFFjFiFyyxxFF,FFFFFFFyxyx/,(cos/),(cos22) jFiF AF2F4F3F1FRxy yiynyyRyxixnxxRxFFFFFFFFFF2121RRyRRRxRyixi

2、RyRxRFFFFFFFFF),cos(,),cos()()(2222jFiF0)()(2222 yixiRyRxRFFFFF 00yixiFF PABCD FBxy0sin, 00cos, 0 AByAxFFFFPF FAPFPFBA21,25 BMFACFBC0coscos, 00sin)(, 0 BABCyBCBAxFFFFFFF0cos, 0 CBMyFFF sin2FFFBABC cot2cosFFFCBM BCFCBFhFMO)(FOABA 2（Nm） AFxOyFyyxxyyOxOOyFxFMMM)()()(FFF)()()()()(21iOnOOOROMMMMMFFFFF)()

3、(xiiyiiROFyFxM FFnOrFrF hcos)(rFhFMnnnOFcos)()()()(rFMMMMnOOrOnOFFFFFFFFFdxFdxFMMMooo)()()(),(FFFF0F FF00FF01F2FABFdFDCF1F2MMMFF FF FF FF F / 2F / 2自由矢量自由矢量F44FF33F1FF1F22FFFdFdFM3111 dFdFM4222 3434,FFFFFF 2134)(MMdFFFdM 0 iM inMMMMM21？aaABCaBCABFBFCFABF02, 0 aFMMA22ABFFMa 22CBBFFFMa M1M2CABDM2CDM1A

4、B FB FA FC FD030cos, 01 MaFMBaFMB231 030sin, 02 aFMMCaFMC212 321 MM0, 0 aFMMAaMFFBA EDCBA FBDBC CFDFMDE FE FDCAE FC EF 045sin, 0 MaFMDaMFFED2 51sin045cossin, 0 ECyFFFaMFC5 EDCBA1F2F1F2FPRMoAMMBCABdABABdMFFFFF3F1F2FF1F2F3niiOOniiRMM11)(FFFFRnixiiyiiniiOOFyFxMM11)()(F22)()(yixiRFFFRyiRRxiRFFFF),cos(,)

5、,cos(jFiF與簡化中心選擇無關(guān)與簡化中心選擇無關(guān)與簡化中心有關(guān)與簡化中心有關(guān)niiOOMM1)(F00ORMF00ORMF00ORMF00ORMF00ORMFOFRFR ROFMd 00ORMFRF 00ORMFFRFRFR00ORMFOFRFR FRFRFR ldxxqdPP0)( lxdxxqxdPPh0)( lldxxqxdxxqh00)()( lqldxxqP0)(2)()(00ldxxqxdxxqhll xlqxq0)( PqPlqxdxlqdxxqPll000021)( 32)()(00ldxxqxdxxqhll 0)(, 0, 0111niiOniyinixiMFFF00O

6、RMF 幾點(diǎn)說明：幾點(diǎn)說明：0)(00FoyxMFF, 0 xF0 cos BAxFF, 0 yF0 sin21 BAyFFGFF , 0FAM 0 sin cos221 lFcFblFlGaFBB6060060 cos, 02 FFFAxx0)(FAM060 sin)(212112 llFlFMlFB, 0 yF060 sin21 FFFFBAykN 75. 0 AxFkN 56. 3 BFkN 261. 0 AyFqq 0224, 002, 00, 0MaaqaGaFMFGaqFFFFBABAyyAxxFqaGFqaGFFBAyAx2143 ,234 ,0 PABCDxy02, 0)(0,

7、 00, 0MaPaFMFFFPFFBABAyyAxxF.,PFPFPFBAyAx 020)(020)(0, 0MaPaFMaPMaFMPFFAyBBAAxxFFPFPFPFBAyAx PABCD02, 0)(02, 0)(02, 0)(MaFaFMMPaaFMPaMaFMBAxCAyBBAFFFPFPFPFBAyAx 0)(, 0)(, 0)(FFFCBAMMM0)(, 0, 0FAyxMFF0)(, 0)(, 0FFBAxMMFFRBAx0223, 0)(0223, 0)(023, 0)(aPMFaMaPMFaMMFaMBCABCAFFFaMFPaMFPaMFCBA3323332,3332

8、 FDECBAaaaACaaaB0)(0FoyMF0)(0)(FFBAMMF3F2F1Fn0 xF0, 00121, 0)(PFFFFFFPMNBNAyNABFN3750,N250 NBNAFFDABCFFFP321 qP0)(, 0)(12lPPebFbaPMNABF0)(, 0)(2bePbFaPMNBAFabePPbalPPe)(21 balPPeP 12abePP)(2 PABCFAFBFCPABFBFA(a)(b)kN3 . 0 AxF解得解得:kN2 . 1 CxF 例題例題8 已知：已知：P=0.4kN, Q=1.5kN, sin=4/5,求：支座求：支座A、C的反力。的反力。A

9、QCBPPABFAxFAyFCxFCyFBxFByFAxFAy) 3(0, 0)2(0sin2cos23cos2, 0)() 1 (0sin2cos2cos2, 0)(QFFFlQlPlFMlQlPlFMCxAxxAyCCyAFF解：解：(1)取整體為研究對象取整體為研究對象解上述方程，得解上述方程，得kN6 . 0,kN2 . 0 CyAyFF(2)取取AB為研究對象為研究對象0coscos2sin, 0)(AyAxBFlPlFMF代入（代入（3）式得）式得 045sin,0045 cos,002522,0GFFFFFFlGlFMEyAyExAxAEF81345 sin825GFGFGFAE

10、yA 02245 cos, 0lFlFlFMEyKDBCF823852GFGFGFDBExK DBFCyFCxF452G2G45, 0 xF045 cos2 GFBx, 0 yF0245 sin2 GGFByN 34745 cos2 GFBxN 847245 sin2 GGFBy0542 CBByFGF, 0 yF, 0 xF053 CBABBxFFFN 340 1 ABFN 660 1 CBF2GBAFBCFByFBxF45節(jié)點(diǎn)構(gòu)造有節(jié)點(diǎn)構(gòu)造有榫接（圖榫接（圖a）焊接（圖焊接（圖b）鉚接（圖鉚接（圖c）整澆（圖整澆（圖d）均可抽象簡化為光滑鉸鏈均可抽象簡化為光滑鉸鏈 FAxFAyFBy012

11、05, 0)(040, 0020, 0ByAByAyyAxxFMFFFFFFFAyFAxA20kNF1F2C10kN10kN10kN10kNAB12345678910111412131516171819212020kNC0, 0020, 012 FFFFyx045sin, 0045cos, 03134 FFFFFFFFAyyAxxkN24kN,16kN,20 ByAyAxFFFkN20, 021 FFkN36kN,21643 FF1FF3F4mn10kNA1234520kNCF6F7F8FAxFAyD0110121, 0)(, 012011, 0)(01045sin, 07867AxCAyDA

12、yyFFFMFFMFFFFFFAxFAyFBy10kN10kN10kN10kNAB12345678910111412131516171819212020kNCkN.42kN,26kN,36876 FFFkN24kN,16kN,20 ByAyAxFFFPEF2F3F4F5FAxFAyF1A2FF603, 0)(0, 00, 0aPaFMPFFFFFNBANBAyyAxxFaaaaaaP1ABECD05 . 0sin5 . 0cos5 . 0, 0)(22aFaFaPMEF0sin, 021 FFFFAyyFAxFAyFNB3/, 3/2, 0PFPFFNBAyAx 3/52PF PF 12EAB

13、CDFAyFAxFECDFDxFDy05 . 1, 0)(05 . 2, 0)(0, 0aqaaFMaqaaFMFFAyBEAAxxFFqaFqaFFEAyAx5 . 25 . 10 045sin5 . 0, 0)(aFaqaMCDFqaFC22 FCBCCAACBFAxFAyMACyFCxFFCxFCy FBFAxFAy FB01, 005 . 012, 0)(0, 0qFFFqFMFFBCyyBCCxxFkN5 . 1, 0,kN5 . 0 CyCxBFFF025 . 11, 0)(01, 00, 0CyAACyAyyCxAxxFqMMMqFFFFFFFmkN4kN,5 . 3, 0 AA

14、yAxMFF若不求若不求C處的力則可處的力則可選整體為研究對象選整體為研究對象DCEFExFEyFDxFDyAHDCGEBFAxFAy) 3(0, 0)2(025002, 0)() 1 (045002, 0)(ExDxxEyDDyEFFFFMFMFFN500,N1000 EyDyFF0650025004, 0)(BAFMFN1000 BFGEBExFEyF0224, 0)(EyExBGFFFMFN1500 ExFN1500 DxFN1000N1500 DyDxFFN500N1500 EyExFFDCEFExFEyFDxFDyAHDCGEBFAxFAyBDAFDyFDxFBxFByFAxFAy解

15、解:(1) 取整體為研究對象取整體為研究對象002, 0)(ByByCFaFMF(2) 取取DEF桿為研究對象桿為研究對象02, 0)(0, 0)( aPaFMaPaFMDxBDyEFF解得：解得：PFPFDxDy2, (3) 取取ADB桿為研究對象桿為研究對象0, 00, 002, 0)(ByDyAyyBxDxAxxDxBxAFFFFFFFFaFaFMF解得：解得：PFPFPFAyAxBx ,aBCDAFEPaaaFCxFCyFBxFByPDFEDxFDyFB例題例題16 求：求：A、D、B的約束反力。的約束反力。aBCDAFEPaaaaBCDAFEaaaAaBCDFEPaaaAaBCDFE

16、aaaPFBxFByFCyFCxBCByFFAyPBxFFAxAB) 3(0, 0)2(0, 0) 1 (02, 0)(CxBxxCyByyByCFFFPFFFaFPaMFPFFCyBy5 . 0 0, 0022, 0)(0, 0BxAxxAyAxBByAyyFFFPaaFaFMPFFFFPFPFPFBxAyAx ,5 . 1,PFCx PPABCDCD022, 0)(0, 00, 0aFaFMMFFFFFFCyCxDDCyDyyCxDxxFPaMPFPFDDyDx 5 . 0CxFCyFPFBxFByFCyFCxBCByFFAyPBxFFAxABPPABCDBCDEDABCDEHFNBFAx

17、FAyFCxFCyFNBDxFDyFFDxFDyFNEH解：解：(1) 取取DE桿為研究對象桿為研究對象kN110, 0322, 0)(DXDxHFqFMMF(2) 取取BDC桿為研究對象桿為研究對象kN3110, 031, 0)( NBNBDxCFFFMF(3) 取整體為研究對象取整體為研究對象0326, 0)(02, 00, 0qFMMMqFFFFFNBAAAyyNBAxxF0,100kNkN,3110 AAyAxMFF解得：解得：例題例題18 已知：已知：q=50kN/m, M=80kNm,求求:A、B的約束反力。的約束反力。02, 00, 021 GGGFFFFFFFByAyyBxAx

18、x 0) 4 2() 2 2() 8 2( 2 5) 2 8 2(, 012GGGGFFMByAF 0m 4m 10)m 2m 4(, 0EBxByCFGFFMF 0m 2m 4m 8, 021GGFMEDFkN 5 . 7,kN 5 .17kN, 5 .77,kN 5 .12 AxBxByEFFFFEFDF2GG1結(jié)論與討論結(jié)論與討論cosFx FjiFyxFFiRFFRRyRRRxRyixiRyRxRFFFFFFFFF),cos(,),cos()()(2222jFiF0iRFF00 yixiFFOABdFMO2)(F)()()()()(21iOnOOOROMMMMMFFFFF力力偶偶矩矩 FdM0 iMiMM nininiiRFF111jiFFyx nixiiyiiniiOOFyFxMM11)()(FROFMd 0)(0 FFFOOiRMM0)(, 0)(, 0)( FFFCBAMMM0)(, 0, 0 FAyxMFF0)(, 0)(, 0 FFBAxMMF力力 系系 名名 稱稱獨(dú)立方程的數(shù)目獨(dú)立方程的數(shù)目平