1目錄第1章緒論2第2章單跨梁的彎曲理論第章桿件的扭轉(zhuǎn)理論15第章力法7第5章位移法28第6章能量法41第7章矩陣法56第9章矩形板的彎曲理論9第10章桿和板的穩(wěn)定性752第1章緒論題）承受總縱彎曲構(gòu)件連續(xù)上甲板，船底板，甲板及船底縱骨，連續(xù)縱桁，龍骨等遠(yuǎn)離中和軸的縱向連續(xù)構(gòu)件（舷側(cè)列板等）承受橫彎曲構(gòu)件甲板強(qiáng)橫梁，船底肋板，肋骨）承受局部彎曲構(gòu)件甲板板，平臺(tái)甲板，船底板，縱骨等）承受局部彎曲和總縱彎曲構(gòu)件甲板，船底板，縱骨，遞縱桁，龍骨等題甲板板縱橫力（總縱彎曲應(yīng)力沿縱向，橫向貨物或上浪水壓力，橫向作用）舷側(cè)外板橫向水壓力等骨架限制力沿中面內(nèi)底板主要承受橫向力貨物重量，骨架限制力沿中面為縱向力艙壁板主要為橫向力如水，貨壓力也有中面力第2章單跨梁的彎曲理論1題設(shè)坐標(biāo)原點(diǎn)在左跨時(shí)與在跨中時(shí)的撓曲線分別為VX與V1X）圖2133323034244246666LLLLLLPXPXPXMXNXVXEIIEIEIEI原點(diǎn)在跨中，323011104466LLPXMXNXVXVEIIEI1111002200LLVVPN）332032266LLPXXMXVXEIIEI圖）333002223666XXXLLPXNQXDXVEIEIEI圖題A33113113126164464164PPPPLPLVVEIEI33512PLEI3333219161929641624PLPLPLVIEIEIB292013366MLLPLVEIIEI2220157316206327PLPLPLEIEIEI2913366MLLPLLEIIEI22201410716206327PLPLPLEIEIEI2221331123336LLPLLVMMEILEI2372430PLIC44475321927682304QLQLQLLVEIEIEI232331116096241668612LQLQLPLQLQLVEIEIIEIEID、和的彎矩圖與剪力圖如圖21、圖2和圖2321圖22圖23圖4圖21圖2圖2323題1）322121062445313120MLQLLLMLQEIIEIEIQL右2）3210173241806QLMLLLMLQEIIEIEI331117131824360612080QLQLEIEI24題，25圖30006NXVXVXEI00VAPN3006XVXAPANI5如圖24，0VLVL由得300200200663LAPLANEILNIPLAPLEIN解出圖2433192PLXVXEILL26圖230012230001212000122122311216,466MXNXVXXEIILVLLNLEIEIMLLLEIIIMLLNLIIXXVXXLL由得解得25題（剪力彎矩圖如）25圖1320233302220033965216184814469186PLMPPRLPPLVAEIEIVLMLPLPLPLVIIIEIVLPLPLPLVLEIIII，圖2516APABMLKL11,0,6632ALALBAK將代入得1632PLPLM6（剪力彎矩圖如）27圖341134224444005240502100512384240100572933844009600LQLQLVAREIEILQLQLVIILQLQLVEIIQLQLIEI圖33312333121120242440100751117242440100300VQLQLQLEILEIEIVQLQLQLLILII（剪力彎矩圖如）28圖2221401112124,0,11,824132121248243,82864AAQABMKLQAALBKQLQLMQLQLRQLVAEI由,代入得圖442433032355238412816384110246246448192648LQLQLMLQLVEIEIIEIVQLLQLILIIQLEILQLQLLMEII726題1MAX2MAX211321213214266240SNEIVSSSSSNDVDXDXDXGGAIVDXVCGAEIAXBXVVFXCXDFXAXBCGAEIIAXBXFXFXCAXDGAQXQXFXFXEIEIVV式中由于1114232343234200002460,2242412242523848SSSSSDBLLQLEIQLALEICALIGAIGAQLALEIQLQLCEIEIQXQLXQXQXQLVXXIIGAEIGALQLQLVEI可得出由得方程組解出A27題先推廣到兩端有位移情形,IIJJ212,IJSEIGAL令32101132216006SIIIIJIIJSJJEIAXBXVCXDAXGADVVCALBLEIVLLALGAALVLBL而由由由8221312IJJIIJALLBLLL解出11210062416426200104261JIIJIJIJJIJIEIMEIVEIBLLILLLEINEIVEIALLLEIMLEIVLEIBALLL令上述結(jié)0IJ果中，即同書中特例28題已知20375225,18,751050KGLCMTCMSCMCM10251007576875KGQHSC1）計(jì)算組合剖面要素形心至球心表面形心至最外板纖維1240950419862TYHECM面積2CM距參考軸CM面積距3CM慣性矩4CM自慣性矩4CM外板1845810002187略球扁鋼24AON387594302231981566045943022539ABC16222460455041166286101198BBECMICCMA275183875174MIN,4555CLLIBESM計(jì)算外力時(shí)面積計(jì)算時(shí)，帶板932118610594433521986TIYECMWCMY32206186101449459422510501740366221086100988,0980IWCMYALUEIXU222212012020176875225098832042412121768752250980158915242415891510501416433532042410501271145032042410503784335QLMXUKGCMQLUKGCMMKGCMWKGCMKGW中中球頭中板固端球頭端（2MAX21416KGCMCM若不計(jì)軸向力影響，則令U0重復(fù)上述計(jì)算222MAX0176875225241050142424433514241416056QLKGWCM球頭中相對(duì)誤差結(jié)論軸向力對(duì)彎曲應(yīng)力的影響可忽略不及計(jì)。結(jié)果是偏安全的。29題220,0IVIVIVEVTVEIVNTVTVVKKIEI式中1,234123413240,0000RRKRKVAKXACHKXASHKXVLEIVNLTVL特征根1034342340ACHKLASHKLEIKLCLPTKASHKLACHKL解得12343,1PPPPATHKLAATHKLAKTKTKTKTVXTHKLXTHKLCXSHKXKPTHKLCHKXSHKXKEIK210題2421,23412341424000,SINCOS000IVIVIVEVTVEVNTVKKEIRRIKRIKVAKXAXAXMEIVMAKEI式中特征方程特征根3433423432234300SINCOS0SINCOSSIN,00COSSINXVLEILTVLAKLAKLLLKAKLAKLMACTGKLATMVKKKKAKEITGKL解得211題圖01320222000644412202436090101880752VLKLEIEILQLMLUUEIEIQLQLMQLU由協(xié)調(diào)條件查附錄圖令A(yù)0B0U1331022213212LLVUVUVULQMVKBEIVU44222104191150663549301193350101210448191154930164800049LUBQLQLEIEIQLI213圖20020001631631,1210720591011112316PLMLXUUMEIEILPLMXUIIULEIPLPL將代入得3133122221332122322480609011109115066354830119335488191154930100086LULLVUVUVLPLMVUEIEIVUPLEIPLI212題1）先計(jì)算剖面參數(shù)12223232261001025044346PIIPBHWCMAYHBHCMFWBH形狀系數(shù)圖MAX42MAX4MAXMAX28A10024008103516168105121655500YYYPKGMWCWPLPKGL）求彈性階段最大承載能力如圖令即解出3UPYWP求極限載荷用機(jī)動(dòng)法此結(jié)構(gòu)達(dá)到極限狀態(tài)時(shí)將出現(xiàn)三個(gè)塑性鉸，其上作用有塑性力矩M如圖由虛功原理圖2422UUPPLPM444240050960500PYUWPKGLLPM213補(bǔ)充題剪切對(duì)彎曲影響補(bǔ)充題，求圖示結(jié)構(gòu)剪切影響下的VX解可直接利用23000066SMXNEIVXVXXXEIIGA13002002223222033663626SSSVVLEIVLMMLLNMMEIILLGAGALXLLXEIVXXEILLGA則邊界條件得214補(bǔ)充題試用靜力法及破壞機(jī)構(gòu)法求右圖示機(jī)構(gòu)的極限載荷P，已知梁的極限彎矩為PM（20分）（1983年華中研究生入學(xué)試題）解1）用靜力法（如圖）由對(duì)稱性知首先固端和中間支座達(dá)到塑性鉸，再加力，當(dāng)PUP作用點(diǎn)處也形成塑性鉸時(shí)結(jié)構(gòu)達(dá)到極限狀態(tài)。即84PUPPUMPLMPL2）用機(jī)動(dòng)法822PPULL215補(bǔ)充題求右圖所示結(jié)構(gòu)的極限載荷其中（1985年哈船工研究生入,3LPQLEI學(xué)試題）解由對(duì)稱性只需考慮一半，用機(jī)動(dòng)法。當(dāng)此連續(xù)梁中任意一個(gè)跨度的兩端及中間發(fā)生三個(gè)塑性鉸時(shí)，梁將達(dá)到極限狀態(tài)?？紤]A、B兩種可能2022240164016LUPPUUPPUAQXDXMLLMQLBLMQL對(duì)解得對(duì)（如圖）取小者為極限載荷為即承受集中載荷P的跨度是破28PUMQL壞。14圖圖15第章桿件的扭轉(zhuǎn)理論31題A由狹長(zhǎng)矩形組合斷面扭轉(zhuǎn)慣性矩公式33334116501020088082643IIJHTCMB33341701251512603JCMC由環(huán)流方程2020222254042440416200830232140416131681643023213168277510TBREDTTMFADSMDSAJAGGDSGTTCMDSTJCM公式材力本題32題對(duì)于A示閉室其扭轉(zhuǎn)慣性矩為423044ATAJTATDSTTT對(duì)于B開口斷面有33143ITJHTAT20032714TBTAMGJATJDST兩者扭轉(zhuǎn)之比為倍）本題易將的積分路徑取為截面外緣使答案為300倍，誤差為10，可用但概念不對(duì)。若采用S為外緣的話，J大，小偏于危險(xiǎn)。33題81222842281SINCOSSIN2828442100309555/230002SIN4TNTBBMPPPBABTBTBTBPMFKGCMABT162451009568SIN88SIN2282SINOS81009568104298OS1002BTLFLFDSBTAGTAGTBTCC弳）34題將剪流對(duì)內(nèi)部任一點(diǎn)取矩112225662322336773784123215636778412312321IIITFRDSFFRDSFRDSFRDSFFRDSFRDSFRDSFRDSFRDSFRDSFRDSFRDSAFFAFM由于I區(qū)與I區(qū)，I區(qū)與I區(qū)扭率相等可得兩補(bǔ)充方程31221122673323372133212213121232212122212231IITFFFFFDSDSDSDSDSGATTGATTTFFDSDSTTFFFFFFFAAAAAFFFMFFFFFF即聯(lián)立（注意到，）13221231212223162300147495214714145TTTTTTMFFAFFFFMMFFADSDSGATTAAGGTMJATJ解得知17第章力法41題0020275262510810251845/20LCMIIIQ令由對(duì)稱性考慮一半噸米對(duì)，節(jié)點(diǎn)列力法方程30100330100102000201201212036240808086324266262426/820902549008171139008421175MLLQLEIIEILLQLMLMLQLIIIIEIEIMQLQLQLTMMT即42題1121223221232212,2632418CQLPQLLMLMLLEIEIEIIQLLLI1將第一跨載荷向支座簡(jiǎn)化M由節(jié)點(diǎn)轉(zhuǎn)軸連續(xù)條件解得182212821668ABQLQLRMLQMQLL2若不計(jì)各跨載荷與尺度的區(qū)別則簡(jiǎn)化為M43題由于折曲連續(xù)梁足夠長(zhǎng)且多跨在A,B周期重復(fù)?？芍髦ё鶖嗝鎻澗厍覟镸對(duì)2節(jié)點(diǎn)列角變形連續(xù)方程333624624MAQAMBQBEIIEIIEII解得2332221121QABQQBAMABB題，4圖21對(duì)，節(jié)點(diǎn)角連續(xù)方程2102001021020017/264341804380346418044101242330/5500182MLLQLMLEIEIEIEILMLLIIIQLQLMLL解得191234023012233404543,IIIIILLLL圖令，由對(duì)稱考慮一半210200210200202012234644547643418043634101242330/5500182MLLQLEIEIEILLLMLLIIIEIEIMQLQLLL（）（）解出45題0020202022000012014412363363642063121133362129451136316LLEIEIMLLMLEIEIEILLEIILKQLM2對(duì)圖剛架對(duì)圖45所示剛架考慮，桿，由對(duì)稱性（）（）均可按右圖示單跨梁計(jì)算。（）由附錄表A6（5）00002041012423307110001821801136755QLQLLLL46題202124222124233MLMLEIEIM節(jié)點(diǎn)平衡為剛節(jié)點(diǎn)，轉(zhuǎn)角唯一（不考慮3桿）222222124212421246,36631321,246MLLLEIKEIEIMEILIEIKKLLEIKL若桿單獨(dú)作用，若桿單獨(dú)作用，兩桿同時(shí)作用，47題已知受有對(duì)稱載荷Q的對(duì)稱彈性固定端單跨梁（），證明相應(yīng)固定系數(shù)與關(guān)EIL系為21EIL36120222111121222IEILMLLQEIILMQIMLQEILEIIML證梁端轉(zhuǎn)角令則相應(yīng)固端彎矩即得或討論1）只要載荷與支撐對(duì)稱，上述結(jié)論總成立2）當(dāng)載荷與支撐不對(duì)稱時(shí)，重復(fù)上述推導(dǎo)可得2111216331112JIJJIJIJIIIJIJIIIJIIJIJIJIJIIIIIORMMII式中外荷不對(duì)稱系數(shù)支撐不對(duì)稱系數(shù)僅當(dāng)即外荷與支撐都對(duì)稱時(shí)有否則會(huì)出現(xiàn)同一個(gè)固定程度為的梁端會(huì)由載荷不對(duì)稱或支撐不對(duì)稱而影響該端的柔度，這與對(duì)梁端的約束一定時(shí)為唯一的前提矛盾，所以適合定義的普遍關(guān)系式是不存在的。48題331211111111131124862331622323,1136ALEILEIPLMLVMLVEILEILEIMPVARPLLPLMPLVEI列出節(jié)點(diǎn)的角變形連續(xù)方程聯(lián)立解出畫彎矩圖見右圖49題）如圖所示剛架提供的121555037672,32PAVPLLMLEIEIPLPLPFL支撐柔度為而由節(jié)點(diǎn)得由卡瓦定理22111122210023333033713177412PPLLPLMAVDSEIPPPSSDSPLSLSDSILLLLSDSEIEIEI）01由對(duì)稱性只需對(duì)，節(jié)點(diǎn)列出方程組求解3011330111131011036246324624122MLLVQLEIILEILLVMLLQLQLIILIIEIIQLQLVAREIL422121136,36,18QLQLQLVEI01聯(lián)立解得MM題3A1384,1192,2192QQALQQQLAKEIAL1212331253331233332221,573103841802855338438438438453841,4851538421648QALQALBQQLLVEIEIQLLQLEIEIILEIQQALQQLAAEIKEIALAL中23334811,4848EIEICPQPKALAL32311143,484448LDPKEIAL23PL8P令P同C圖48EI6EI6548551,238448321614812132496367QQALQQALQQLAPP3EPLPL令48EI6311122227681313247687712227MMEIKALLL327PLLF令768EI6EIP303033022192611922226192QALQLGPAPQQPAEILKXAKEIAILKXALKAEIAL同即411題0,V支柱處可簡(jiǎn)化為剛性固定約束僅考慮右半邊板架24330004044001481111642164481611162484869296122PPEIKEIALEILLLILLUEIP00101116616087404507881108520852029292232PLPLMPLPPNPPMAX0333300010416089615289912LVVVLPLPLPLEIEIEI412題54001002620001132158331010251857,12YQSINLYSINSIN0707SIN1L42ALMIICMLLLBLIIQQALQKGEKGCMVCC中設(shè)求中縱桁跨中及端部彎曲應(yīng)力及解因主向梁兩端簡(jiǎn)支受均布載荷故其形狀可設(shè)為251112211I112121236221130020836441100143264164100208307070014321004110707400411,0013023841058331IICIIKEIAL按對(duì)稱跨中求352202200304400003112122100411102830013023168004111010185700411122120728,0813,0774316811007280304283KGCMQLQQQLLLLIUILLIALIUUQVUCMK中22200522100512316810M081351101012I51121083310U31681007745148124I51241083310QLLQLKGCMHITQLLQLKGCM端中413補(bǔ)充題寫出下列構(gòu)件的邊界條件（15分）261）1V0A00PEIVVLLEIM解2）1122V000EIVMVLEIVLML解3設(shè)X0,B時(shí)兩端剛性固定；Y0,A時(shí)兩端自由支持200X0,BY0,AYWW解時(shí)時(shí)XW4已知X0,B為剛性固定邊Y0邊也為剛性固定邊YA為完全自由邊0X0,BY00WYW解時(shí)X時(shí)222200YXYAWWYX時(shí)414題圖示簡(jiǎn)單板架設(shè)受有均布載荷Q主向梁與交叉構(gòu)件兩端簡(jiǎn)支在剛性支座上，試分析兩向梁的尺寸應(yīng)保持何種關(guān)系，才能確保交叉構(gòu)件對(duì)主向梁有支持作用解少節(jié)點(diǎn)板架兩向梁實(shí)際承受載荷如圖，為簡(jiǎn)單起見都取為均布載荷。由27對(duì)稱性由節(jié)點(diǎn)撓度相等12RR331332QL5L384EI48EIL115RL972I162I11QQ32WQALQLLQBLQLL主交12使之相等令3515R112944816201QLLLILI解出節(jié)點(diǎn)反力DR式中交叉構(gòu)件與主向梁的相對(duì)剛度，且由節(jié)點(diǎn)反力將隨的增加即交叉構(gòu)件剛性的增加）而增加。當(dāng)55481224QLLQLLMAX時(shí)R這時(shí)交叉構(gòu)件對(duì)主向梁的作用相當(dāng)于一個(gè)剛性支座當(dāng)表示交叉構(gòu)件的存在不僅不支持351I101294IRL3且符合荷載彎曲條件T12CM222226606560406/24124AAMQBKGCTT22222606560812/1212BQBKGCMT244433631106560002843007322101238412AGBGBCMTETEB已知中面力20188/KGCM220036311126012188103082/12221012TUBBUEET221224241632230656009259022424065600957186612121065093600663843210121/612/6024ABAQBMUKGQBUKGGBUFUCMDWTCM09021885638/024AAMKGC018661889655/024BAKGCMW與9（A）比較可見，中面拉力使板彎曲略有改善，如撓度減小，彎曲應(yīng)力也略有減少，但合成結(jié)果應(yīng)力還是增加了。921）當(dāng)板條梁僅受橫荷重時(shí)的最大撓度442MAX63555580103384384210/12QLD009171外加中面力對(duì)彎曲要素的影響必須考慮（本題不存在兩種中面力復(fù)合的情況）3）2200226605808008006/405888AAMQLUTT上下452800348/1148KGCM93已知T06CM，L60CM，Q1KG/CM2，36342/12100639560109112ETDCMU）判斷剛性考慮僅受橫荷重時(shí)的424MAX363515060091384/1238421006/12QLUET427CM，必須考慮彎曲中面力。MAX1/427/06715T2）計(jì)算超靜定中面力（取K05）64421121006003109116005ETUUQLK由圖97查曲線A得U314LOG10249由線性查值法00353313313021301660213020204FFFF0222031020153020201916239560422460/14224/06704/TDKGLTKGCM440551600204087038438439560QLFUD中222MAX0026616070401912137/8068QLUKGCMT94設(shè)滿足解，代入微方程,SINMXXYFYA0,XA724442,QXYXXYYD設(shè)關(guān)于的常微分方程MFY（1）24,2SINIVMMMXQXYFYFYFYAAAD為定現(xiàn)將也展成相應(yīng)的三角級(jí)數(shù)，其中MFY,QXY,SINMXQXYQYA02,SINAMMXQYQXYDXA本題可看成（0的極限情景）0,BQXYQ000COSCOS22LIMSINLIM22LIMSINSINCMCMCMXAAQYDXAACCAAAA將代入方程（1）右邊比較得2,SINSINMCMXQXYAAA2422SINIVMMMMCFYFYFYAADA特解（2）4SINMPACFYD特征方程成對(duì)雙重根42420MMMSSSAAA齊次解為MMMMMYFYACHYBSHCYCHYDYSHYAAAA73由于撓曲面關(guān)于X軸對(duì)稱，所以通解中關(guān)于Y的奇函數(shù)必然為0。（）0MBC通解MMMMFYACHYDYSHYFYAAA其中可按處即求解。,MAD/2YB22/0Y0MMFYFY即式中2202MMMMMUUCHFYUACDCHSHMBUA解出222MMMMMUFYADTHUCH2222MMMMMFYUFYFYTHCHYYSHYFYUUAAACHCH2222MMMFYUUMMYSHYTHCYCHUAAACH將（2）中代入得,SINMXXYFYA342SINMAMCFYDA344SIN,22SIN22MMMMCUUPAMMXAXYYSHYTHCYCHUDAAACH95已知A板中心垂直于X軸斷面應(yīng)力2222660540600/577818XQAKGCMT結(jié)論按荷形彎曲計(jì)算的結(jié)果彎曲要素偏大，所以偏于安全。原因是按荷形彎曲計(jì)算時(shí)，忽略了短邊的影響，按（長(zhǎng)邊A）/（短邊B）計(jì)算。表中A/B所對(duì)應(yīng)數(shù)值，即表示按荷形彎曲計(jì)算結(jié)果。96設(shè)顯然滿足幾何邊界條件21,SINSIN4MXYXYAB0,00XA時(shí)，但Y時(shí)，00YB時(shí)，令取一項(xiàng)SINSI4XYAAB則SICOS,SINCOS44YXYXYXYAABAB222222SINSI,SISIN4XYYXYXAYAAB2COSCOS4XXYB222222220V212ABDUDXDYXYXYXY2222222202222SINSIN21444COSCOSSINSIN44ABDXYAUAABABABXYXYDXYABAB75222222322222212111444416212812048DABABAUAABAUABAB024202423232223EI,ISINEI21IA21024B16128122048ABBYBAVDYYADYBBVDBAABEIA002,SINSI44A2ABABXYUQXYDXDYQADXDYABAB322232A162128121024AB40VUDBAUABEIAAQAB解出452223422234222342102416212812784162896228784SINSI4,162896228QABADBAUABEIAQABBAAIAXYGABABXYDBAEIA76第10章桿和板的穩(wěn)定性101題（A）取板寬350MIN,MIN,7570CM55ELBB（但計(jì)算中A的帶板取75）LI（屬大柔度桿）35011021665136100LIA（KG/CM2）22262/10/1361067CRE（直接由查圖時(shí)只能準(zhǔn)確到100KG/CM2，KG/CM2）1100CR（B）面積（）IA2CM對(duì)參考軸的靜矩IAZ3CM慣性矩2IAZ4CM自身慣性矩0I4CM帶板140007023112立板10110（51）10621103翌板6565（1105）65105265312156501282510766313054ABC12071722408211021CMECMABICECA77取代板寬，200MIN,40CM555ELLBB求面積A時(shí)取50EB扶強(qiáng)材兩端約束可視為簡(jiǎn)支（屬于小柔度桿）20036132636113100LIA）2U37013710372537263723728372979（1）2211223823MLMLVVUULEILEI式中21121212822LLEITTUUTUEIEIL22122LTUUUEI虛設(shè)彈性支座反力（2）220MTVTVRLL12簡(jiǎn)化關(guān)于,V的聯(lián)立方程組22112230123LLMUUVLEIEIMT失穩(wěn)時(shí)，V不能同時(shí)為零，故其系數(shù)行列式為零。即21122341201TLUULEI化簡(jiǎn)后穩(wěn)定方程為222231322XUTGXTGTGUTGU由圖解法或數(shù)值解法可得其最小根（見下說明）2MIN1705XU2221075291EEIEITLL說明80如下圖，最小根必然在區(qū)間（）內(nèi)，即（157，222）2XU,2再由數(shù)值列表由線性內(nèi)差法求解1的對(duì)應(yīng)X值為12Y2170517017009511170488100209511XU105題1）計(jì)算有關(guān)參數(shù)12210,05VVEB12125,336N525V查圖，跨縱骨作為剛支座上連續(xù)壓桿的歐拉應(yīng)力2262021012506164/6405250EIKGCMAL2）求橫梁對(duì)縱骨的支持剛度X161701705171018TG76966740657137222XTG7320273979732021Y0951110012102568144644336210500050101964/500EIBKKGCMB橫梁臨界剛度4465331010036421012505673/250JNCRJXEIKXKGCML可見0CRCRK需要進(jìn)行非彈性修正4）逐步近似法確定，令CR00654JX由線性內(nèi)差法計(jì)算221002050205000654005982100/0070000598CRYKGCM4K令2CRYDKBT即22/121YKETBU2222624007009112121626410YTKE128TCM107題2/CRKGCM（表F1）00CR查（）JXJX160008888547902920002400213180007