船舶结构力学课程习题集答案

11目录第1章绪论2第2章单跨梁的弯曲理论2第章杆件的扭转理论15第章力法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75第1章绪论题）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）承受横弯曲构件甲板强横梁，船底肋板，肋骨）承受局部弯曲构件甲板板，平台甲板，船底板，22纵骨等）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1题设坐标原点在左跨时与在跨中时的挠曲线分别为VX与V1X）图213332304246666LLLLLLPXPXPXMXNVEIEIEIEI原点在跨中，32301114LLXXVXII11022LLVVPN）3320266LLXMXVXEIEI图）3330023XXXLLPQDIII图题A331111264644PLLVEIEI352L331916926424PLPLLVIIEIB036MLPLVIEI2221571603LLLEIII33291336MLPLLEII222014076LLLEIII221331236LPLLVMEII2740LEC4531968QLLLLVIII232331096462LQLPLQLQLVEIEIEIID、和的弯矩图与剪力图如图21、图22和1图图图图23图2144图22图2323题1）322121064530MLQLMLQEIIEIL右2）321017486QLLLQIII33112360280LLEIEI24题，25图00NXVXI00VAPN36APE如图24，0VL由得5530020020663LAPANEILPLLIN解出图243192PLXVXEIL26图230012230012122311126,46MXNVXEILVLEILLILNNEXXVXLL由得解得25题（剪力弯矩图如）25图1320233022096521841696PLMPRLVAEIEILLPPLIVLLLVI，1APABMLK图251,0,632ALL将代入得163PL（剪力弯矩图如）7图6634124405201538207936LQLVAREIILQLVILQLEEI图333121204407510VQLQLQLILIEILLLEL（剪力弯矩图如）28图222140124,0,811438,86AAQABMKLQLQLQLMLRQVAEI由,代入得图442433023552816810192648LLQLMLQLIEIEIVLLQLILQLLMEIE26题9910101MAX2AX1321213214620SNEIVSSSSSDVDXGACAEIAXBVFCXDFXABCGAIFXFDGQQXFFEIEIV式中由于114232342342060,1425238SSSSSDBLQILAICALEGAGALIQLLCIEIXQXQLVXGAIALQLLI可得出由得方程组解出A111127题先推广到两端有位移情形,IJ21,IJSEIGAL令31011321606SIIIIJIIJSJJIAXBVCXDAXVALBEIVLLALGALL而由由由2231IJJIIJALLBLL解出1121062416420146JIIJIJIJJIJIEIMEIVIBLLLLINIVIALLLEIMEILIBLL令上述结0IJ果中，即同书中特例28题已知203752,18,7515KGLCMTCSCMC0568QHS面积2CM距参考轴C面积距3CM惯性矩4C自惯性矩4CM外板1845810002187略1212球扁钢24AON3875943022232119815660459430222539ABC116622246045504186109BBECMICCMA75837IN,LLIBESC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形心124095186TYHM至最外板纤维32186594TIYECMWCY320618074068198,9IWCALUEIXU22212012017685098347501589441504632704583QLMXUKGCMKGCWMMKG中中球头中板固端球头端（2AX2146KGCMC若不计轴向力影响，则令U0重复上述计算222MAX017685401443652QLKGWCM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。29题1313220,0IVIIVEVTVNTTKKEI式中1,234132400RKRVAXCHASXVLEINLTVL特征根34234ACHKSILCHKLPKASHKLCL解得12343,1PPATHKLTHKLTTTVXXCSXPTHKLHKEI210题2421,34214200SINCOS0IVIVIEVTEVNTVKKRRIKVAKXAXMEIEI式中特征方程特征根34323432234300SINCOS0INCOSIN,0COSSIXVLITVLAKLAKLAKLLMTGTMVKKKKEITGL解得211题图01414320220064414691875VLKLEIILQLMUEIQLQL由协调条件查附录图令ALVUVUULQMVKBEI44222104950634901350118169LUBQLQLEIEIL213图200163,2107591316PLMLXUEIIEULIPPL将代入得313312222132122480690564809589318LULLVUVULPLMVUEIILIPL212题1）先计算剖面参数15152323261025046PIPBHWCMAYHBCFWBH形状系数图MAX42MAX4AMAX281021035651165YYYPKGMWCWPLKGL）求弹性阶段最大承载能力如图令即解出3UPYPW求极限载荷用机动法此结构达到极限状态时将出现三个塑性铰，其上作用有塑性力矩M如图由虚功原理图242UUPLP405960PYUKGLLP213补充题剪切对弯曲影响补充题，求图示结构剪切影响下的VX解可直接利用230006SMXNEIVXXEIGA1616002022322366SSSVVLEIVLMMLNMEILGAGALXLXEIVXIL则边界条件得214补充题试用静力法及破坏机构法求右图示机构的极限载荷P，已知梁的极限弯矩为（20分）（1983年华中研究生入学试题）PM解1）用静力法（如图）由对称性知首先固端和中间支座达到塑性铰，再加力，当PU作用点处也形成塑性铰时结构达到极限状态。即84PUPUMLML2）用机动法82L215补充题求右图所示结构的极限载荷其中（1985年,3PQLEI哈船工研究生入学试题）解由对称性只需考虑一半，用机动法。当此连续梁中任意一个跨度的两端及中间发生三个塑性铰时，梁将达到极限状态。考虑A、B两种可能2022401640LUPPUPAQXDMLLBQL对解得对（如图）取小者为极限载荷为即承受集中载28PUMQL荷P的跨度是破坏。1717图图第章杆件的扭转理论31题A由狭长矩形组合断面扭转惯性矩公式333341650128026IJHTCM1818B3334170251260JCMC由环流方程2022225404441608316370TBREDTTMFADSMSAJGGDSGTACMDSTJ公式材力本题32题对于A示闭室其扭转惯性矩为42304ATAJTDST对于B开口断面有314IJHTAT203714TBAMGJATDST两者扭转之比为倍）本题易将的积分路径取为截面外缘使答案为30倍，误差为10，可用但概念不对。若采用S为外缘的话，J大，小偏于危险。33题81222421SINCOSSIN84410395/2SITNTBMPPBATTBPFKGCMT245109568SIN8SIN22O109648OS0BTLFLFDSBTAGTTTCC弪）191934题将剪流对内部任一点取矩11222566377378412325661231IIITFRDSFRDSFRSFRSRSFDDAFFFM由于I区与II区，II区与III区扭率相等可得两补充方程3122167327132122211231213IITFFFFFDSSDSDSGATTGATTTFFTTFFFFAAAFFMF即联立（注意到，）132212322231630047495141445TTTTTTMFAFFDSGATTAGGMJAT解得知2020第章力法41题0020275618145/0LCMIIQ令由对称性考虑一半吨米对，节点列力法方程3010031020002120210364886426/954873MLQLEIILLMLLQLEIIEIQLLMLTM即42题112122312322,63418CQLPLMLLEIIEILLMQ将第一跨载荷向支座简化由节点转轴连续条件解得212821668ABQLLRLLL2若不计各跨载荷与尺度的区别则简化为M43题由于折曲连续梁足够长且多跨在A,B周期重复。可知各支座断面弯矩且为M对2节点列角变形连续方程33364624AQQEIIEI解得212123221121QABQQBAMA题，4图对，节点角连续方程2102001021020017/6341843143/5082LLQLMLEIIEIIMLLLLLQ解得12340230123405,IIILL图令，由对称考虑一半22222102002102002020123464571836343/502MLLQLEIIEILLLMLLIEILLMQ（）（）解出45题00202020001201413636426323394561LLEIIMLLLIIELLIKQLM2对图刚架对图5所示刚架考虑，杆，由对称性（）（）均可按右图示单跨梁计算。（）由附录表A（5）000020412788375QLLLLL46题2124223MLLEIIM节点平衡为刚节点，转角唯一（不考虑3杆）23232222124246,3663131,MLLEIKEIIMILEEIKLLI若杆单独作用，若杆单独作用，两杆同时作用，47题已知受有对称载荷Q的对称弹性固定端单跨梁（），证明相应固EIL定系数与关系为21EIL361202211122IEILMLQEILMQLEILIL证梁端转角令则相应固端弯矩即得或讨论1）只要载荷与支撑对称，上述结论总成立2）当载荷与支撑不对称时，重复上述推导可得2424121633112JIJJIJIJIIIIJIIJIIJIJIIIIIORMMIIII式中外荷不对称系数支撑不对称系数仅当即外荷与支撑都对称时有否则会出现同一个固定程度为的梁端会由载荷不对称或支撑不对称而影响该端的柔度，这与对梁端的约束一定时为唯一的前提矛盾，所以适合定义的普遍关系式是不存在的。48题331211111131124863332,6ALEILIPLMLVLVIIEIMARLLPLLVEI列出节点的角变形连续方程联立解出画弯矩图见右图49题）如图所示刚架提供的1215503762,32PAVLMLEIIPLPFL支撑柔度为而由节点得由卡瓦定理252511221002333071174PPLLPLMAVDSEIPSLDSLSDIEIEI）1由对称性只需对，节点列出方程组求解3013131016246242MLVQLEIEIMLQLLIEVARIL422126,36,8QLQLQLVEI01联立解得M题3A184,192,92QQALQQLKEI123312533123332,57108408345841,48526QALLBQLLVEIIQLLIILEIQQALQLAIKILL中262633481,48EIICPQKALL32311143,484LDPKEIAL23PL8令同C图8EI6I6548551,2384321648124967QQALQQALLPPEPLL令EI6I3111222768347687MEIKALLL3PLLF令EII300330221961962QAQGPAQPAEILKXKEIAKXLAIL同即411题0,V支柱处可简化为刚性固定约束仅考虑右半边板架2727330044001481624862892961PEIKEIALILLLLUI00106874581522923PLPLMPLNPMAX033330001468961528912LVVLPLPLPLEIIEI412题5400100262011321835857,YQSINLYSINI07SIN1L4ALMIICMLLBLIIQQALKGEKGCMVCC中设求中纵桁跨中及端部弯曲应力及解因主向梁两端简支受均布载荷故其形状可设为28281121I122236130283641460837014320174,51ICIKEIAL按对称跨中求35220200304400311212218685741278,8,6833KGCMQLQQLLLLIUILIALIUUQVCMK中22052210513681M8310I5U368744I541830QLQLKGCMHITLQKGC端中413补充题写出下列构件的边界条件（15分）1）29291V0A00PEIVVLIM解2）1122V000IVVLEIVLM解3设X0,B时两端刚性固定；Y0,A时两端自由支持200X,BY,AWW解时时4已知X0,B为刚性固定边Y0边也为刚性固定边YA为完全自由边0X,BY0WY解时时22220WYXAW时414题图示简单板架设受有均布载荷Q主向梁与交叉构件两端简支在刚性支座上，试分析两向梁的尺寸应保持何种关系，才能确保交叉构件对主向梁有支持作用解少节点板架两向梁实际承受载荷如图，为简单起见都取为均布载荷。由对称性由节点挠度相等12R30303312QL5RL84EIIL97I6I1132WQALQBLL主交12使之相等令355R194816201LLILI解出节点反力DR式中交叉构件与主向梁的相对刚度，且由节点反力将随的增加即交叉构件刚性的增加）而增加。当554812QLLLMAX时R这时交叉构件对主向梁的作用相当于一个刚性支座当表示交叉构件的35I01294IRL时即时存在不仅不支持主向梁，反而加重其负担，使主向梁在承受外载荷以外还要受到向下的节点反作用力这是很不利的。只有当时，主向梁才受到交叉构件的支持。3I1LIL31313232第5章位移法51题图44，01012QLM15021L023M，20124LIE2014LIE203023/LILI203032/4LEILIM对于节点2，列平衡方程即2130213213M代入求解方程组，有，解得1548030200QLLEILILE02302154EIQL所以200012120084124153ILQLMLQLLEI20002121061855LLMLLI图。由对称性知道045231），102QL1501L023M2），20124LIEM2014LIE33330002332246EIIEIMLLL3）对2节点列平衡方程231M即，解得002216605IQLIL20215QLEI4）求（其余按对称求得）2123,M200012120084124153IQLLQLLLEI2000212106185ILLLLI，其余，2321M4321M3421323M52题由对称性只要考虑一半，如左半边1）固端力（查附表A4），2120015MQLQL2210051MQLQL532342）转角对应弯矩（根据公式5,5），01224EIML0124EIL52000252IIILLL，0023234EIILL003223IILL430003432EIIEILLL3）对于节点2，3列出平衡方程即4215230M图51（单位）20QL343432432451125MM则有，得000233200220008415EIIELLLIQLLLLL302301456QEIL4）3222001212000421574654EIQLMLQLQLLI322200210008615417ILQLLLLEI32200250045ILMLQLLI3322000023004211613575455EIQLEIQLMQLQLLILI332200003200861411ILILLLLEILEI其余由对称性可知（各差一负号），6512M，5621M52542343；弯矩图如图51434353题（），其余固端弯矩14250128PL218MPL都为0，411EIML41IL522EIL524EIL，633263，1212ILL112IMLL，3234EI2234EI3535由1、2、3节点的平衡条件即421523360M142142253135626MM11222323348480EIEIPLLLLIEIPLLLIIILLL解得，21764PEI2516PLEI23564PLEI2141270LMLLLI241738640IPLPLLEI2363245501463EIPLMLPLMLI263277ILLLLI225455068168EIPLLPLLI222335049747ILELMLPLLII21253701654PLL25228416EILPLLI弯矩图如图52363654题已知，1203LM206LLM24039L，4IC1II2408I，0210QQLL40Q42001369LLQ矩三角（）1）求固端弯矩，20ML1205QL23M00493615LL023512L2）转角弯矩，0012124EIIMLL0021124IILL，0023234EIIMLL0032234IILL图52（单位）QL3737，024283EIML0422IL3）对1、2、3节点列平衡方程即214230M0012001300028154796631EIIQLLLILLLLII解得，2001439738QLQLEIEI，2200297637LQLII220039381474QLQLEIEI4）求出节点弯矩210029148738017MQLL23006624LL2400931717QLL24002536MLL弯矩图如图53。55题由对称性只考虑一半；节点号12杆件号IJ1221230/IJI43IJL11IJK43图53（单位）0QL3838IJC1（1/2）对称IJK43/2IJ11/2IJ8/113/11IJN1/20/IJMQL1/101/1500/IJIJML4/1658/1651/55IJ41/3301/551/55所以，012431QLM021345QLM0235L56题1图54令010120120,5IILL节点号012杆件号IJ011012210/IJI11IJL115IJK12/3IJC13/4IJ11/2IJK3/2IJ2/31/3IJN1/200/IJMQL1/101/15000/IJIJML1/452/451/45IJ11/901/451/450由表格解出012MQL12L22210M2图55令，103I012I，LL，0Q10QQL012L节点号012杆件号IJ011012210/IJI31IJL11IJK31IJC11IJ31IJK4IJ3/41/4IJN1/21/22/IJMQL1/121/1211/1925/1922/IJIJMQL5/5125/2565/7685/15362/IJML00931006380063800228由表格解出，20193QL21012638MQL2108QL若将图55中的中间支座去掉，用位移法解之，可有422569183LLEIQ解得，3327051496LLEII422753QLQLII，120ML234Q，21NL222304NQL57题计算如表所示节点号1234杆件号IJ1221232432420/IJI238IJL1223IJK215/118/3IJC3/43/41IJ3/245/448/3IJ198/685297/15071056/2055IJN001/20/IJMQL02/15033021/50/IJIJML0091530624116273008136IJ010487062411627305013658题1）不计杆的轴向变形，由对称性知，4、5节点可视为刚性固定端2,230032QQLL3400618QQLL,2/15M323009LL24/QQ3计算由下表进行，21812039ML2076Q，2334058L251，24309MQL23017MQL，其它均可由对称条件得出。5273939节点号12345杆件号IJ1812212523323443520/IJI11116612IJL6113333IJK1/6111/3224IJC1/2111111IJ1/12111/3224IJK13/1210/3IJ1/1312/130301061/32/3IJN1/21/21/21/21/21/22/IJMQL0000030450450450045009003018009015000346041540207701500306032/IJIJMLQ00537010730035802146010730017900041004960024800179003580071500358000640012800043002560012800022404000005000590003000022000430008500043000080001600005000310001600003000030000500011000060000100000000020000120/IJMQL000390003900786003410112705181051810415900170图54A图54B414159题任一点I的不平衡力矩为（I1，2，,H,I,J,N1SI1,I1）012IISQLM所以任一中间节点的分配弯矩与传导弯矩均为0。IJMIJJINM任一杆端力矩IJIJIJIJ0IJIJISJIJIJSIJMNMN对两端，由于只吸收传导弯矩0,IN0IJMIJIJIJIJ两端所以对于每个节都有杆端力矩IJIJ说明对图54B所示载荷由于也能使，也可以看作两端刚固的0IM单跨梁。4242第6章能量法61题1）方法一虚位移法考虑B,C所示单位载荷平衡系统，分别给予A示的虚变形MXDEI外力虚功为IJ1W虚应变能为L001VDEIX0011LIIILIIIRMXIDEJIIJJIJI1BI36EI2MMCELLLL由虚功原理得WVIIJJ123IL2）方法二虚力法（单位虚力法）梁弯曲应力MXYIEIIJIXXL10ML4343给以虚变化虚应力为IM1IMXYI虚余功WI虚余能（真实应变）（虚应力）VDXMYXYZEII2201LAD/1/LIIJXLLDXI132IIJLQME同理给以虚变化，可得（将换为）JJ0IMIJ32IJJLI3）方法三矩阵法（柔度法）设，虚IIJJ,P,IJPMP/1IIIJJMXYYXXLCPIIL式中（不妨称为物理矩阵以便与刚度法中几何矩阵1,YCIL对应）B虚应力IJMCP实应变11DCP虚余功TTIIJJWM虚余能VDD444411TTPCDPDPCDDP于虚力原理考虑到虚力的任意性。得WV1ATPDP式中柔度矩阵（以上推导具有普遍意义）ACD对本题220111LXXXLLYXLDDIEILEILL/3/61/21/LLLII由展开得AP1/2/3IIJJMLEI62题方法一单位位移法，/JIUL/JIUL设，则1I/1IUL20/LIJIJIIJEEAEATULDUDXULL同理，令可得1J1/JJITULDLJIUL即可记为1IIJJEALIJIJPK为刚度矩阵。K方法二矩阵虚位移法4545设TIJIJPTIJIJU1/IJIIJJULB式中几何矩阵1BLIJDB设虚位移，虚应变TIJIJUIJB外力虚功TIJIJIJIJWPP虚应变能TVDDTIJIJBDTIJIJDIJIJK由得WVIJIJP式中刚度矩阵TBDD对拉压杆元详细见方法一。11LEAKEAXLL方法三矩阵虚力法设，IIJJTPIIJJUD1IJIIJJTCPA式中物理矩阵（指联系杆端力与应力的系数矩阵）C虚应力11IJDPIJCP设虚力，则IIJJTP1IJD4646虚余功TTIJIJIJIJWP虚余能VDD1TIJIJCDP1TIJIJPDIJIJA式中柔度矩阵1TCDD对拉压杆1LLKXEAEAIJIJP即1IIJJUTLEA讨论比较方法二、三。结论，IJIJPKIJIJAP若与的逆矩阵存在（遗憾的是并非总是存在），则，实际上是一个柔度矩阵，实际上是一个刚度矩阵1163题1）63如图所示0设12COSNXVXAL显然满足处的0,L变形约束条件VL变形能20LEIVVDX2201COSLNINXADLL421NIL4747力函数（对称）2PVCLPVC12OSNAL由，所以。即0NVNVA4221COSNEILAPL所以，344NNCLLEI344122COS1COSNPLNXVXILL2）64如图所示0设01SINXVXAL2222402011SI2LNNVLALEIXEIVDALLALA01INCPUCAPL由0V得，2/ALAPC所以，20/L由，得NVA所以，4SI2EILCPL342SINNPLCALEI3241ISINACXVXLILL48483）65如图所示令02VXALX所以，20LEIVD20236LAXI/24005192LLQUDQLXDQAL由得所以，VA345419AEILL768EI25768QLVXXI4）66所示如图，0设，231A123VAX0LEIVVDX1204LID213LL231/2/LLQVXAX312758ALL由得10V3127/4EILALQ由得2A24165/6ILLL解上述两式得22173849QLEIA301745067QLQLVXXXEII64题如图所示4949设1SINXVXAL/4/2204LLEIIVDVDX4242/4/221104SINSINLLXEIAXIADLL4213LI1100SIN2/LLXQVXDADQLAL由得1V413LEIL所以，44507832QQLAIEI41SINLXUXI65题如图所示设12SINXVXAL2220LVEIVDXA242311SIN2NNIAL其中，3LAEI433121212SINSINNNVIAALL505022001SIN2LLXQVXDADL114CO2NNALQA所以，41NQL取前两项得，41123312VEIIAALL4212332VEIIAALL由得104133IIQLL由得2VA421334EIEILALL即4121708493QILAE解得412078QLIA43018SIN012SINXXQLVXLEI中点挠度4762QI66题取12SIN,SINXXVXAVBLL20022200422422SINCOSLSLLSNNSNNSNNGAEIVDDGAXXABDLLLLEILBLALLA4,NNLVA10011SIICOCOSLLLLNQVXDXQDXQNL51511COS,1COSNNLLAQBQN由445520NNVLLEIEI为奇数得由22331COSNNNSQLQLBAGAA为奇数得12UXX42553341SINSIN,3SQLQLXEILLN67题1）图69对于等断面轴向力沿梁长不变时，复杂弯曲方程为0IVETQ取能满足梁段全部边界条件SINXVXAL00,0,0LIVLVVETQVDX处有420SINSINSILNNXNEIATALLLLL积分420COLNNXIQLL即4425201COS1/2NNQALNEILNLEIUT为偶数为奇数式中今已知U1LUI452SIN1,351/NXQLLVXEI44521092/NLLQLEIIUN取一项5252准确解为4440551710925823838LQLQLQLVFEIEIEI误差仅为046结论1）引进2CRNTL单跨简支压杆临界力254,38LUEI2）取一项，中点挠度表达式可写成如下讨论的形式440513838CRCRQLTLQLVEIEIT失稳的压力时式中当T为拉力时取正号（此时相当一缩小系数，随T而）1当T为压力时取负号（此时相当一放大系数，随T而）12）图610弹性基础梁平衡方程为0IVEKVQ00LIVEKVQDX取代入上式SINXAL40ISINSI0LNNXXNXAEIKAQDLLLL由于的随意性有式中积分为0，即N41COS2NNLLLEIAKAQN44541COS1/2NLQQLNIKLEINKIL为奇数由44LUKUIEIL得代入得535344521NQLAUEIN454SI1,351NXQLLVXNIKEIL今取一项，且令U1，求中点挠度445078221LQLQLEII准确值4401086252LQLUEIKI误差为85误差较大，若多取几项，如取二项则误差更大，交错级数的和小于首项，即按级数法只能收敛到略小于精确解的一个值，此矛盾是由2L于是近似值。068题2201LMXDAREI梁支0234334216216LLRQXDILLQLREEILI由最小功原理解出0VR528QL54544345223880175QLRLVEIIQ中69题由对称性可知，对称断面处剪力为零，转角，静不定内力和可00TM最小功原理求出21020/SIN1COSQSMOASRRTRAB段段001OB段段段段最小功原理0022221001/SIN1COSSRVMSDEIQMQRRTRRDEI22200SIN1COSCSVRTRRRTEI分别得2001641324MRQTR解得0538274QRST表达式正确由得极值点在点,该处极值为1MS0T10M5555由得20MS02785,0629QRTGT极值为2221538SIN7451COS061RQRQR区间端点B处222538SIN7451079MRQQRQRMAX012,79BBMR发生在支撑处610题由左右对称,对陈断面01上无剪力。有垂向静力平衡条件0SIN2PQRD解得/4QPR5656任意断面弯矩为200000PRSIN1COS1COS2RININ21,COSMSTQRDRT有最小功原理确定T0和M0200PR1CSINSI0VRQRDMEI即22P0Q2001PR1COSINSI1COS0TRQRRDTI0CSS0MDMD即220PROCINSICOS0RTRQD得2004/TQT与图中假设方向相反0PR48M2PRPRR1COSIN4SIR85757第7章矩阵法71题解由CH2/24题/26图计算结果32121XLXLXV，2121LLXLLV21216JILXLYV2264EDLLXB,13222222204321313441231TELXYXLKDDLLXXLLLEIEIILLL对称72题解如图示离散为3个节点，2个单元585823232121