上海交大与哈工程通用船舶结构力学课本课后习题答案

1目录第1章绪论2第2章单跨梁的弯曲理论第章杆件的扭转理论15第章力法7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9第10章杆和板的稳定性752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，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2133323034244246666LLLLLLPXPXPXMXNXVXEIIEIEIEI原点在跨中，323011104466LLPXMXNXVXVEIIEI1111002200LLVVPN）332032266LLPXXMXVXEIIEI图）333002223666XXXLLPXNQXDXVEIEIEI图题A33113113126164464164PPPPLPLVVEIEI33512PLEI3333219161929641624PLPLPLVIEIEIB292013366MLLPLVEIIEI2220157316206327PLPLPLEIEIEI2913366MLLPLLEIIEI22201410716206327PLPLPLEIEIEI2221331123336LLPLLVMMEILEI2372430PLIC44475321927682304QLQLQLLVEIEIEI232331116096241668612LQLQLPLQLQLVEIEIIEIEID、和的弯矩图与剪力图如图21、图2和图2321图22图23图4图21图2图2323题1）322121062445313120MLQLLLMLQEIIEIEIQL右2）3210173241806QLMLLLMLQEIIEIEI331117131824360612080QLQLEIEI24题，25图30006NXVXVXEI00VAPN3006XVXAPANI5如图24，0VLVL由得300200200663LAPLANEILNIPLAPLEIN解出图2433192PLXVXEILL26图230012230001212000122122311216,466MXNXVXXEIILVLLNLEIEIMLLLEIIIMLLNLIIXXVXXLL由得解得25题（剪力弯矩图如）25图1320233302220033965216184814469186PLMPPRLPPLVAEIEIVLMLPLPLPLVIIIEIVLPLPLPLVLEIIII，图2516APABMLKL11,0,6632ALALBAK将代入得1632PLPLM6（剪力弯矩图如）27图341134224444005240502100512384240100572933844009600LQLQLVAREIEILQLQLVIILQLQLVEIIQLQLIEI图33312333121120242440100751117242440100300VQLQLQLEILEIEIVQLQLQLLILII（剪力弯矩图如）28图2221401112124,0,11,824132121248243,82864AAQABMKLQAALBKQLQLMQLQLRQLVAEI由,代入得图442433032355238412816384110246246448192648LQLQLMLQLVEIEIIEIVQLLQLILIIQLEILQLQLLMEII726题1MAX2MAX211321213214266240SNEIVSSSSSNDVDXDXDXGGAIVDXVCGAEIAXBXVVFXCXDFXAXBCGAEIIAXBXFXFXCAXDGAQXQXFXFXEIEIVV式中由于1114232343234200002460,2242412242523848SSSSSDBLLQLEIQLALEICALIGAIGAQLALEIQLQLCEIEIQXQLXQXQXQLVXXIIGAEIGALQLQLVEI可得出由得方程组解出A27题先推广到两端有位移情形,IIJJ212,IJSEIGAL令32101132216006SIIIIJIIJSJJEIAXBXVCXDAXGADVVCALBLEIVLLALGAALVLBL而由由由8221312IJJIIJALLBLLL解出11210062416426200104261JIIJIJIJJIJIEIMEIVEIBLLILLLEINEIVEIALLLEIMLEIVLEIBALLL令上述结0IJ果中，即同书中特例28题已知20375225,18,751050KGLCMTCMSCMCM10251007576875KGQHSC1）计算组合剖面要素形心至球心表面形心至最外板纤维1240950419862TYHECM面积2CM距参考轴CM面积距3CM惯性矩4CM自惯性矩4CM外板1845810002187略球扁钢24AON387594302231981566045943022539ABC16222460455041166286101198BBECMICCMA275183875174MIN,4555CLLIBESM计算外力时面积计算时，带板932118610594433521986TIYECMWCMY32206186101449459422510501740366221086100988,0980IWCMYALUEIXU222212012020176875225098832042412121768752250980158915242415891510501416433532042410501271145032042410503784335QLMXUKGCMQLUKGCMMKGCMWKGCMKGW中中球头中板固端球头端（2MAX21416KGCMCM若不计轴向力影响，则令U0重复上述计算222MAX0176875225241050142424433514241416056QLKGWCM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220,0IVIVIVEVTVEIVNTVTVVKKIEI式中1,234123413240,0000RRKRKVAKXACHKXASHKXVLEIVNLTVL特征根1034342340ACHKLASHKLEIKLCLPTKASHKLACHKL解得12343,1PPPPATHKLAATHKLAKTKTKTKTVXTHKLXTHKLCXSHKXKPTHKLCHKXSHKXKEIK210题2421,23412341424000,SINCOS000IVIVIVEVTVEVNTVKKEIRRIKRIKVAKXAXAXMEIVMAKEI式中特征方程特征根3433423432234300SINCOS0SINCOSSIN,00COSSINXVLEILTVLAKLAKLLLKAKLAKLMACTGKLATMVKKKKAKEITGKL解得211题图01320222000644412202436090101880752VLKLEIEILQLMLUUEIEIQLQLMQLU由协调条件查附录图令A0B0U1331022213212LLVUVUVULQMVKBEIVU44222104191150663549301193350101210448191154930164800049LUBQLQLEIEIQLI213图20020001631631,1210720591011112316PLMLXUUMEIEILPLMXUIIULEIPLPL将代入得3133122221332122322480609011109115066354830119335488191154930100086LULLVUVUVLPLMVUEIEIVUPLEIPLI212题1）先计算剖面参数12223232261001025044346PIIPBHWCMAYHBHCMFWBH形状系数图MAX42MAX4MAXMAX28A10024008103516168105121655500YYYPKGMWCWPLPKGL）求弹性阶段最大承载能力如图令即解出3UPYWP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2422UUPPLPM444240050960500PYUWPKGLLPM213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23000066SMXNEIVXVXXXEIIGA13002002223222033663626SSSVVLEIVLMMLLNMMEIILLGAGALXLLXEIVXXEILLGA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PM（20分）（1983年华中研究生入学试题）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PUP作用点处也形成塑性铰时结构达到极限状态。即84PUPPUMPLMPL2）用机动法822PPULL215补充题求右图所示结构的极限载荷其中（1985年哈船工研究生入,3LPQLEI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2022240164016LUPPUUPPUAQXDXMLLMQLBLMQL对解得对（如图）取小者为极限载荷为即承受集中载荷P的跨度是破28PUMQL坏。14图图15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33334116501020088082643IIJHTCMB33341701251512603JCMC由环流方程2020222254042440416200830232140416131681643023213168277510TBREDTTMFADSMDSAJAGGDSGTTCMDSTJCM公式材力本题32题对于A示闭室其扭转惯性矩为423044ATAJTATDSTTT对于B开口断面有33143ITJHTAT20032714TBTAMGJATJDST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300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81222842281SINCOSSIN2828442100309555/230002SIN4TNTBBMPPPBABTBTBTBPMFKGCMABT162451009568SIN88SIN2282SINOS81009568104298OS1002BTLFLFDSBTAGTAGTBTCC弪）34题将剪流对内部任一点取矩112225662322336773784123215636778412312321IIITFRDSFFRDSFRDSFRDSFFRDSFRDSFRDSFRDSFRDSFRDSFRDSFRDSAFFAFM由于I区与I区，I区与I区扭率相等可得两补充方程31221122673323372133212213121232212122212231IITFFFFFDSDSDSDSDSGATTGATTTFFDSDSTTFFFFFFFAAAAAFFFMFFFFFF即联立（注意到，）13221231212223162300147495214714145TTTTTTMFFAFFFFMMFFADSDSGATTAAGGTMJATJ解得知17第章力法41题0020275262510810251845/20LCMIIIQ令由对称性考虑一半吨米对，节点列力法方程30100330100102000201201212036240808086324266262426/820902549008171139008421175MLLQLEIIEILLQLMLMLQLIIIIEIEIMQLQLQLTMMT即42题1121223221232212,2632418CQLPQLLMLMLLEIEIEIIQLLLI1将第一跨载荷向支座简化M由节点转轴连续条件解得182212821668ABQLQLRMLQMQLL2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333624624MAQAMBQBEIIEIIEII解得2332221121QABQQBAMABB题，4图21对，节点角连续方程2102001021020017/264341804380346418044101242330/5500182MLLQLMLEIEIEIEILMLLIIIQLQLMLL解得191234023012233404543,IIIIILLLL图令，由对称考虑一半210200210200202012234644547643418043634101242330/5500182MLLQLEIEIEILLLMLLIIIEIEIMQLQLLL（）（）解出45题0020202022000012014412363363642063121133362129451136316LLEIEIMLLMLEIEIEILLEIILKQLM2对图刚架对图45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A6（5）00002041012423307110001821801136755QLQLLLL46题202124222124233MLMLEIEIM节点平衡为刚节点，转角唯一（不考虑3杆）222222124212421246,36631321,246MLLLEIKEIEIMEILIEIKKLLEIKL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（），证明相应固定系数与关EIL系为21EIL36120222111121222IEILMLLQEIILMQIMLQEILEIIML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2111216331112JIJJIJIJIIIJIJIIIJIIJIJIJIJIIIIIORMMII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331211111111131124862331622323,1136ALEILEIPLMLVMLVEILEILEIMPVARPLLPLMPLVEI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121555037672,32PAVPLLMLEIEIPLPLPFL支撑柔度为而由节点得由卡瓦定理22111122210023333033713177412PPLLPLMAVDSEIPPPSSDSPLSLSDSILLLLSDSEIEIEI）01由对称性只需对，节点列出方程组求解3011330111131011036246324624122MLLVQLEIILEILLVMLLQLQLIILIIEIIQLQLVAREIL422121136,36,18QLQLQLVEI01联立解得MM题3A1384,1192,2192QQALQQQLAKEIAL1212331253331233332221,573103841802855338438438438453841,4851538421648QALQALBQQLLVEIEIQLLQLEIEIILEIQQALQQLAAEIKEIALAL中23334811,4848EIEICPQPKALAL32311143,484448LDPKEIAL23PL8P令P同C图48EI6EI6548551,238448321614812132496367QQALQQALQQLAPP3EPLPL令48EI6311122227681313247687712227MMEIKALLL327PLLF令768EI6EIP303033022192611922226192QALQLGPAPQQPAEILKXAKEIAILKXALKAEIAL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24330004044001481111642164481611162484869296122PPEIKEIALEILLLILLUEIP00101116616087404507881108520852029292232PLPLMPLPPNPPMAX0333300010416089615289912LVVVLPLPLPLEIEIEI412题54001002620001132158331010251857,12YQSINLYSINSIN0707SIN1L42ALMIICMLLLBLIIQQALQKGEKGCMVCC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251112211I112121236221130020836441100143264164100208307070014321004110707400411,0013023841058331IICIIKEIAL按对称跨中求352202200304400003112122100411102830013023168004111010185700411122120728,0813,0774316811007280304283KGCMQLQQQLLLLIUILLIALIUUQVUCMK中22200522100512316810M081351101012I51121083310U31681007745148124I51241083310QLLQLKGCMHITQLLQLKGCM端中413补充题写出下列构件的边界条件（15分）261）1V0A00PEIVVLLEIM解2）1122V000EIVMVLEIVLML解3设X0,B时两端刚性固定；Y0,A时两端自由支持200X0,BY0,AYWW解时时XW4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0X0,BY00WYW解时X时222200YXYAWWYX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27对称性由节点挠度相等12RR331332QL5L384EI48EIL115RL972I162I11QQ32WQALQLLQBLQLL主交12使之相等令3515R112944816201QLLLILI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55481224QLLQLLMAX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存在不仅不支持351I101294IRL3且符合荷载弯曲条件T12CM222226606560406/24124AAMQBKGCTT22222606560812/1212BQBKGCMT244433631106560002843007322101238412AGBGBCMTETEB已知中面力20188/KGCM220036311126012188103082/12221012TUBBUEET221224241632230656009259022424065600957186612121065093600663843210121/612/6024ABAQBMUKGQBUKGGBUFUCMDWTCM09021885638/024AAMKGC018661889655/024BAKGCMW与9（A）比较可见，中面拉力使板弯曲略有改善，如挠度减小，弯曲应力也略有减少，但合成结果应力还是增加了。921）当板条梁仅受横荷重时的最大挠度442MAX63555580103384384210/12QLD009171外加中面力对弯曲要素的影响必须考虑（本题不存在两种中面力复合的情况）3）2200226605808008006/405888AAMQLUTT上下452800348/1148KGCM93已知T06CM，L60CM，Q1KG/CM2，36342/12100639560109112ETDCMU）判断刚性考虑仅受横荷重时的424MAX363515060091384/1238421006/12QLUET427CM，必须考虑弯曲中面力。MAX1/427/06715T2）计算超静定中面力（取K05）64421121006003109116005ETUUQLK由图97查曲线A得U314LOG10249由线性查值法00353313313021301660213020204FFFF0222031020153020201916239560422460/14224/06704/TDKGLTKGCM440551600204087038438439560QLFUD中222MAX0026616070401912137/8068QLUKGCMT94设满足解，代入微方程,SINMXXYFYA0,XA724442,QXYXXYYD设关于的常微分方程MFY（1）24,2SINIVMMMXQXYFYFYFYAAAD为定现将也展成相应的三角级数，其中MFY,QXY,SINMXQXYQYA02,SINAMMXQYQXYDXA本题可看成（0的极限情景）0,BQXYQ000COSCOS22LIMSINLIM22LIMSINSINCMCMCMXAAQYDXAACCAAAA将代入方程（1）右边比较得2,SINSINMCMXQXYAAA2422SINIVMMMMCFYFYFYAADA特解（2）4SINMPACFYD特征方程成对双重根42420MMMSSSAAA齐次解为MMMMMYFYACHYBSHCYCHYDYSHYAAAA73由于挠曲面关于X轴对称，所以通解中关于Y的奇函数必然为0。（）0MBC通解MMMMFYACHYDYSHYFYAAA其中可按处即求解。,MAD/2YB22/0Y0MMFYFY即式中2202MMMMMUUCHFYUACDCHSHMBUA解出222MMMMMUFYADTHUCH2222MMMMMFYUFYFYTHCHYYSHYFYUUAAACHCH2222MMMFYUUMMYSHYTHCYCHUAAACH将（2）中代入得,SINMXXYFYA342SINMAMCFYDA344SIN,22SIN22MMMMCUUPAMMXAXYYSHYTHCYCHUDAAACH95已知A板中心垂直于X轴断面应力2222660540600/577818XQAKGCMT结论按荷形弯曲计算的结果弯曲要素偏大，所以偏于安全。原因是按荷形弯曲计算时，忽略了短边的影响，按（长边A）/（短边B）计算。表中A/B所对应数值，即表示按荷形弯曲计算结果。96设显然满足几何边界条件21,SINSIN4MXYXYAB0,00XA时，但Y时，00YB时，令取一项SINSI4XYAAB则SICOS,SINCOS44YXYXYXYAABAB222222SINSI,SISIN4XYYXYXAYAAB2COSCOS4XXYB222222220V212ABDUDXDYXYXYXY2222222202222SINSIN21444COSCOSSINSIN44ABDXYAUAABABABXYXYDXYABAB75222222322222212111444416212812048DABABAUAABAUABAB024202423232223EI,ISINEI21IA21024B16128122048ABBYBAVDYYADYBBVDBAABEIA002,SINSI44A2ABABXYUQXYDXDYQADXDYABAB322232A162128121024AB40VUDBAUABEIAAQAB解出452223422234222342102416212812784162896228784SINSI4,162896228QABADBAUABEIAQABBAAIAXYGABABXYDBAEIA76第10章杆和板的稳定性101题（A）取板宽350MIN,MIN,7570CM55ELBB（但计算中A的带板取75）LI（属大柔度杆）35011021665136100LIA（KG/CM2）22262/10/1361067CRE（直接由查图时只能准确到100KG/CM2，KG/CM2）1100CR（B）面积（）IA2CM对参考轴的静矩IAZ3CM惯性矩2IAZ4CM自身惯性矩0I4CM带板140007023112立板10110（51）10621103翌板6565（1105）65105265312156501282510766313054ABC12071722408211021CMECMABICECA77取代板宽，200MIN,40CM555ELLBB求面积A时取50EB扶强材两端约束可视为简支（属于小柔度杆）20036132636113100LIA）2U37013710372537263723728372979（1）2211223823MLMLVVUULEILEI式中21121212822LLEITTUUTUEIEIL22122LTUUUEI虚设弹性支座反力（2）220MTVTVRLL12简化关于,V的联立方程组22112230123LLMUUVLEIEIMT失稳时，V不能同时为零，故其系数行列式为零。即21122341201TLUULEI化简后稳定方程为222231322XUTGXTGTGUTGU由图解法或数值解法可得其最小根（见下说明）2MIN1705XU2221075291EEIEITLL说明80如下图，最小根必然在区间（）内，即（157，222）2XU,2再由数值列表由线性内差法求解1的对应X值为12Y2170517017009511170488100209511XU105题1）计算有关参数12210,05VVEB12125,336N525V查图，跨纵骨作为刚支座上连续压杆的欧拉应力2262021012506164/6405250EIKGCMAL2）求横梁对纵骨的支持刚度X161701705171018TG76966740657137222XTG7320273979732021Y0951110012102568144644336210500050101964/500EIBKKGCMB横梁临界刚度4465331010036421012505673/250JNCRJXEIKXKGCML可见0CRCRK需要进行非弹性修正4）逐步近似法确定，令CR00654JX由线性内差法计算221002050205000654005982100/0070000598CRYKGCM4K令2CRYDKBT即22/121YKETBU2222624007009112121626410YTKE128TCM107题2/CRKGCM（表F1）00CR查（）JXJX160008888547902920002400213180007