船舶结构力学习题集答案(交大版)

1目录第1章绪论.......................................................................................2第2章单跨梁的弯曲理论................................................................2第３章杆件的扭转理论................................................................15第４章力法.....................................................................................17第5章位移法...................................................................................28第6章能量法...................................................................................41第7章矩阵法...................................................................................56第9章矩形板的弯曲理论..............................................................69第10章杆和板的稳定性................................................................752第1章绪论１．１题１）承受总纵弯曲构件：连续上甲板，船底板，甲板及船底纵骨，连续纵桁，龙骨等远离中和轴的纵向连续构件（舷侧列板等）２）承受横弯曲构件：甲板强横梁，船底肋板，肋骨３）承受局部弯曲构件：甲板板，平台甲板，船底板，纵骨等４）承受局部弯曲和总纵弯曲构件：甲板，船底板，纵骨，递纵桁，龙骨等１．２题甲板板：纵横力（总纵弯曲应力沿纵向，横向货物或上浪水压力，横向作用）舷侧外板：横向水压力等骨架限制力沿中面内底板：主要承受横向力货物重量，骨架限制力沿中面为纵向力舱壁板：主要为横向力如水，货压力也有中面力第2章单跨梁的弯曲理论２.1题设坐标原点在左跨时与在跨中时的挠曲线分别为v(x)与v(1x)１）图2.133323034243()()()424()26666llllllpxpxpxMxNxvxEIEIEIEIEI原点在跨中：3230111104()4()266llpxMxNxvxvEIEIEI，'11'11()0()022(0)0(0)2llvvpvN２）33203()32.2()266llpxNxMxvxxEIEIEI图３）333002()22.3()666xxxllpxNxqxdxvxxEIEIEI图２.２题a)33111311131(3)(2)616444641624pppplplvvvEIEI=3512plEI333321911()61929641624plplplVEIEIEI3b)2'292(0)(1)3366MlMlPlvEIEIEI=2220.157316206327PlPlPlEIEIEI2291()(1)3366MlMlPllEIEIEI=2220.1410716206327PlPlPlEIEIEI2222133311121333363llpllvmmEIlEI=2372430plEIc)44475321927682304qlqlqllvEIEIEI23233'11116(0)962416683612lqlqlplqlqlvEIEIEIEIEId)2.1图、2.2图和2.3图的弯矩图与剪力图如图2.1、图2.2和图2.3图2.14图2.2图2.32.3题1）32212120624452313120MlqlllMlqqEIEIEIEIqlM右2）32101732418026qlMlllMllqEIEIEIEI=3311117131824360612080qlqlEIEI2.4题2.5图3000()6NxvxvxEI，00vApN300()6xvxApxANEI5如图2.4，0vlvl由得300200200060263lAplANEIlNEIplAplEIpN解出3333()1922plxxvxEIll图2.42.6图2300122300012120001221223121212260,42026622MxNxvxxEIEIvlvlMlNlEIEIMlllEIEIEIMlNlNlEIEIxxvxxll由得解得2.5题2.5图：（剪力弯矩图如２．５）132023330222002332396522161848144069186plMppRpllpplvAREIEIvlMlplplplvEIEIEIEIvMlplplplvlEIEIEIEI16ApabbMAlKl，图2.5111,0,6632AlalbAK将代入得：16312plplM62.7图：（剪力弯矩图如２．６）341113422244440.052405021005112384240100572933844009600lqlqlvAREIEIlqlqlvAREIEIlqlqlvEIEIqlqlEIEI图２．６3331233312111202424401007511117242440100300vvqlqlqlEIlEIEIvvqlqlqllEIlEIEI2.8图（剪力弯矩图如２．７）2221401112124,,0,11,82411118243212121248243,82864AAQabMAKlQqaalbAKqlqlMqlqlqlRqlvAREI由,代入得图２．７442433032355238412816384111(0)246246448192()6488lqlqlMlqlvEIEIEIEIvqlMlqlEIlEIEIqlEIlqlqllMEIEI72.6题.1max2max2113212132142.()()62()()62()()242(0)sNEIvsssssNdvdxdxdxGGANEIvdxvCGAGAEIaxbxvvvfxcxdfxaxbCGAEIEIaxbxfxfxcaxdGAGAqxqxfxfxEIEIvv式中由于11142323432342(0)00()()00242602,224()241222425()23848sssssdbvlvlqlEIqlalEIcalEIGAEIGAqlalEIqlqlcEIEIqxqlxqxqxqlvxxEIEIGAEIGAlqlqlvEIGA可得出由得方程组：解出：a=2.7.题先推广到两端有位移,,,iijj情形：212,ijsEIGAl令321011322162(0)(0)()62()2siiiijiijsjjEIaxbxvcxdaxGAvdvvcalblEIvllalGAalvlbl而由由由82213121ijjiijallblll解出1121(0)(0)62416642162(0)(0)1()(0)()()4261jiijijijjijiEIMEIvEIbllEIlllEINEIvEIallNlNEIMlEIvlEIballl令上述结0ij果中，即同书中特例2.8题已知：20375225,1.8,751050kglcmtcmscmcm1025100.7576.875kgqhscm面积2cm距参考轴cm面积距3cm惯性矩4cm自惯性矩4cm外板1.84581000(21.87)略球扁钢ON24a38.759430.22232119.815.6604.59430.22253.9ABC=11662224604.55.04116628610119.8BBecmICcmAA2751.838.75174min,4555AcmllIbescm计算外力时面积计算时，带板1）.计算组合剖面要素：形心至球心表面1240.95.0419.862tyhecm形心至最外板纤维9321186105.94433.5219.86tIyecmwcmy32206186101449.45.9422510501740.3662221086100.988,()0.980IwcmyAluEIxuu222212012020176.8752250.988320424.1212176.8752250.980158915)242415891510501416433.53204241050127114503204241050378433.5qlMxukgcmqlMukgcmMkgcmwMkgcmwMkgw中中球头中板固端球头端（2max21416kgcmcm若不计轴向力影响，则令u=0重复上述计算：222max0176.875225241050142424433.5142414160.56%1424qlkgwcm球头中相对误差：结论：轴向力对弯曲应力的影响可忽略不及计。结果是偏安全的。2.9.题220,0,0IVIVIVEIvTvEIvNTvTTvVvKVkEIEI式中1,234123413240,0(0)0(0)00()0()()()rrkrkvAAkxAchkxAshkxAAvvAAvlEIvlNlTvl特征根：10343342340AchklAshklEIkAs