工程力学教学第一部分

作业：NO:11、图没画：2、图中的符号与公式不一致；3、无计算过程：公式＝答案；4、脱离体，没完全脱离开来。作业中的主要问题：1－5计算图中三个力分别在ｘ、ｙ、ｚ轴上的投影。已知F1=2kN,F2=1kN,F3=3kN。题1－5图55√25435F1F2F3Oxyz11111341.2,1.6,055xyzFFkNFFkNF22222230.424,5240.566,5210.7072xyzFFkNFFkNFFkN331330,3xyzFFFFkN解：1-7求图示的两个力对A的矩之和。512200N390N6m3m4mA1252003390639041313AM＝2160N.m1-8求图示的三个力对A点和B点的矩之和。101.53045AMkNmB2m1.5mA30kN30kN10kN1m101.53045BMkNm1－9钢缆ＡＢ中的张力FT=10kN.。写出该张力FT对ｘ、ｙ、z轴的矩及该力对Ｏ点的矩（大小和方向）。题1－9图FTxFTyFTz解2222014.142oxyzMMMMkNm=++==,221cos(),cos(,),cos(,)333OOOMxMyMz=-==-102.35732102.357321049.42832TxTyTzFkNFkNFkNz2mFTxyO4m1m1mAB202429.43320249.43310224.713xTyTzyTxzTxMFFkNmkNmMFkNmkNmMFkNmkNml1BFAOxyzl3l21-10求力F对O点的矩的矢量表达式122212322221233222123xyzlFFllllFFllllFFlll322222123311222123210xzyzzxyllFMFllllllFMFllllMFlFlFxFyFz3231222222123123ollFllFMijllllll1-11。工人启闭闸门，为了省力，常常用一根杆子插入手轮中，并在杆的一端C施力FC能将闸门开启，若不借用杆子直接在A,B处施加力偶（F,F'),问F应为多大才能开启闸门？FF'FC=200NCAB解：(/2),cFABFACAB150FN1-13FOαFPBA30°FPABFAXFAFAYFBFBFACFPABOFCFCABFACOFOXABFAOFBFOYBFBαFP1AFP2αFP1ABFAYFAXFAYFBαFPABCFPABFAXFPFPABCDFPCDFPABCFBFAxFAyFCxFCyFCxFCyFDFAxFAyFBFD2—6图示混凝土管搁置在倾角为30°的斜面上，用撑架支承，水泥管子重量FQ=5kN。设A、B、C处均为铰接，且AD=DB，而AB垂直于斜面。撑架自重及D、E处摩擦不计，求杆AC及铰B的约束力。ABCDEFQ45°30°xABDFAFB0cos451.77ixBDFFFkN===åo0cos451.77iyACDFFFkN===åosin302.5DQFFkN==oFQFDy2-11图示结构已知M=1.5kNm,a=0.3m。求铰A和C的约束反力。BBCA2aaMFAFC3cos45AFaM1.52.363cos4530.32/2ACMFFkNa2—12滑道摇杆机构受两力偶作用，在图示位置平衡。已知OO1＝OA=0.2m,M1＝200Ｎ·ｍ，求另一力偶矩M2及O，O1两处的约束力（摩擦不计）。AOO130°M1M21cos30,1155AAFOAMFkN==o122,400AFOAMMNm==?OM1AFAO1AO130°M2FAo43F1F2FM221.50.53-60OM112432221.5055FFFFM40FN3-7ABl/2l/2FqFAABlqFBq2ABlq1FAFBFAFB+22ABFqlFF3BqlF6AqlF12123BqlqqFl12126AqlqqFlFPMAB3.5m0.5mFAxFAyFBFAxF2F13m2.5mCB2.5mqAFAxFAyMA3-83-940BFkN0AxF20AyFkN4AxFkN17AyFkN43AMkNmF=4kNM=6kNmABC4m4mFBFAyFAxMABC:0,1.5CBMFkN0,0ixAxFF0,2.5iyAyFFkN整体：0,10AiAMMkNm3－12整体：DC:3－131mDABC4m4mFPFQ1m4m4m0,810,6.25CDDMFFFkN轮右0,164790,105ADBBMFFFFFkN轮左轮右－0,0,51.25iyDBAAFFFFFFFkN轮左轮右－解：起重机左右轮压分别为F轮左＝10kN和F轮右＝50kNFBFDFADABCF轮右F轮左FAFDFB1m0.5m2m1mABFq0CDq3-14FDFAxFAyFBDC:整体：2.5DFkN=∑MCi=00.3AxFkN=∑Fix=03.54BFkN=∑MAi=00.538AyFkN=-∑MBi=03－15CAq=5kN/mFAFABFCxFCy4.5m4.5m1.5mBCAq=5kN/mFAFB4.522.5ABFFqkN33.833.84.54.50ABCxABCyFkNFFkNFqq3-16ACE:311.67FkN=整体：114.58FkN=3m4m4m3mAB4kNE7kNC6kN4m4m3m3mH123FAFBF33m3mA4kNEC6kN4m12F1F3EF22结点E:28.75FkN=-8BFkN=∑MAi=09AFkN=∑MBi=0∑MCi=0∑Fix=0∑Fiy=00.8m0.6m1.5mAB3.4kNCD3-17FCBC3.4kNFCDFAFD3.41.50,8.50.6DiAMFkN0,11.9iyDFFkN1.70,3.47.2250.81.50,6.3751.7iyCDixCBCDFFkNFFFkNFBDBFBCFBAFBDDFDFADFDC求反力：1.7m0,/0.610.6270,0.88.501ixBABCiyBDBAFFFkNFFFkN1.50,6.3751.7ixADCDFFFkN3－186m4m4m4m3mABGE4kNCD2kN1231329FkN=-329FkN=-2BxFkN=III-I右侧:∑MCi=0∑MDi=0∑Fix=0FByFBxFA整体：83ByFkN=∑Fyi=043AFkN=∑MBi=0GF34kNF2C23DFByFBxF12209FkN=∑Fyi=03－193m4m4m4m3m3mABGE25kNCDH123FGE25kNF1DH123FGF325kNF2H23F183.3FkN=333.3FkN=-241.67FkN=-IIIIIII-I:II-II:∑MGi=0∑MEi=0∑Fiy=0ACBP4-7xzyFAFBFCBCFF=4002APFN==∑Myi=0∑Mxi=0200BCFFN==∑Fiz=04-9xyzFzFxFyMyMxMz30.yMkNm=-32.xMkNm=5xFkN=-∑Myi=0∑Mxi=0∑Fix=04ByFkN=-∑Fiy=08zFkN=∑Fiz=020.zMkNm=∑Mzi=0q=2kN/mF1=5kNF2=4kN4m2m4m0.51.50.20.21.10.8xyo4-14求形心0.751.50.50.60.81.51.11.50.3/20.7031.50.50.81.51.50.3/2cxm11.50.50.250.81.51.250.31.51.020.87911.50.50.81.50.31.52Cym31:(0.5,1.5,1.0),1326xyzVm32II:(2.5,1.0,1.0),32212,xyzVm33III:(4.5,1.0,0.4),120.81.6xyzVm0.562.5124.51.62.056121.6iiCixVxmV1.561.0121.01.61.156121.6iiCiyVymV1.061.0120.41.61.956121.6iiCizVzmV解：xyz34121o0.82IIIIII4-15