理论力学(自制)

1、计算题一平面任意力系（考固定端约束）BCCAACBFAxFAyMACyFCxFFCxFCy FBFAxFAy FB01, 005 . 012, 0)(0, 0 qFFFqFMFFBCyyBCCxxFkN5 . 1, 0,kN5 . 0 CyCxBFFF025 . 11, 0)(01, 00, 0 CyAACyAyyCxAxxFqMMMqFFFFFFFmkN4kN,5 . 3, 0 AAyAxMFF若不求若不求C处的力则可处的力则可选整体为研究对象选整体为研究对象PFBxFByFCyFCxBCByFFAyPBxFFAxAB) 3(0, 0)2(0, 0) 1 (02, 0)( CxBxxCyBy

2、yByCFFFPFFFaFPaMFPFFCyBy5 . 0 0, 0022, 0)(0, 0 BxAxxAyAxBByAyyFFFPaaFaFMPFFFFPFPFPFBxAyAx ,5 . 1,PFCx PPABCDCD022, 0)(0, 00, 0 aFaFMMFFFFFFCyCxDDCyDyyCxDxxFPaMPFPFDDyDx 5 . 0CxFCyFPFBxFByFCyFCxBCByFFAyPBxFFAxABPPABCDBCDEDABCDEHFNBFAxFAyFCxFCyFNBDxFDyFFDxFDyFNEH解：解：(1) 取取DE杆为研究对象杆为研究对象kN110, 0322, 0)

3、( DXDxHFqFMMF(2) 取取BDC杆为研究对象杆为研究对象kN3110, 031, 0)( NBNBDxCFFFMF(3) 取整体为研究对象取整体为研究对象0326, 0)(02, 00, 0 qFMMMqFFFFFNBAAAyyNBAxxF0,100kNkN,3110 AAyAxMFF解得：解得：例题例题3 已知：已知：q=50kN/m, M=80kNm,求求:A、B的约束反力。的约束反力。计算题二运动学（速度合成、加速度合成）reavvv coscos0aelvvvD retanaaaaa enatasincosaaa )sincos(200e lareavvv sinsiner

4、vvv aeaanratrarnrteaaaaa nrea cossinaaa 3222asin cot)sin cos( sin1RvaRvaa 计算题三运动学（刚体的定轴转动+平动基点法求速度）计算题三运动学（刚体的定轴转动+平动基点法求速度）计算题三运动学（刚体的定轴转动+平动基点法求速度）AO1OB3203vAvB0 OAvAAB03rOBvABB AO1OB20215ratB nAatAanBatBanAatAanBAatBAanBAtBAtAnABnBaaaaaa t200203,rraratAnA 2022rABaABnBA 201223rBOvaBnB nBAnAtAnBtBa

5、aaaa 60cos30cos30cos60cos计算题四动量矩定理（ 动量矩定理+质心运动定理）lglmgml23,2312 OyOxFmglmFlm 20224, 0mgFFOyOx OABC260sin12160sin60cos2)()(lFmlJmaFmgFmaFFACcyAeyCxAex 2, 0laatCAnCAnCAtCAAC aaaa230sin,30coslaaaaACyAcx BAC60aAaCAaAACB.132,1318,1332galgmgFAA 计算题五动能定理（ 速度合成+质心运动定理）AO30CxAO30COvAO30COvOv iiWTT122222212121CCOmvJmvT mgllmgWii41)30sin1 (2 lvlvvvvvvOOCAOCAOC214160cos2222222 OOvmlmlmvT41612222 mglvmlmlmvTTOO41416122212 0)60cos2( lvmmvOOlvO81 lg2934 AO30CxAO30COvAO30COvOv iiWTT122222212121CCOmvJmvT mgllmgWii41)30sin1 (2 lvlvvvvvvOOCAOCAOC214160cos2222222 OOvmlmlmvT41612222 mglvmlm