电力出版社材料力学课后习题答案

解作轴力图如下400KN340KN70KN130KNA400270340KNFF/3F/3F/3BF2F/3F/3213FF2FC3F2FLL/2LFFF/LFDF2FFN题71图2F2FFFF2FFE2FF/LF/L23A2M2M4M12MABC1MDQ10KN/M12FAFBFCXFCYFAAFN2CDQ10KN/M1FN2FN1DN10404104236422ICMFKN120,5/2407,IXFFFKN3111636410/354,115010FAMPA3222640710/317115010FAMPA解求反力,FAFB40KN由结点D的平衡取一半如图分析23B1M1M1M1MABECDQ3KN/M123AF1ECDQ3KN/MF2F1CF310,3315/1135AMFKN310,2191IXFFFKN230,1/2135IYFFFKN31161326336135101598501013510225600101911038250010FMPAAMPAMPA解去掉1杆的约束如图,用约束力F1代之,由由结点C的平衡3MAX6MIN14103502010410FMPAA368102673010BCMPA解AFNF10KN,B轴力图如图,AB17KN9KN16KNCD11B22B0FFA24BKN81724（B36241020012010CDMPA3617102138010ABMPA361210403010BCMPA3618104540010ABMPAMAX45BCMPA（C）,12KN50CMABCC50CM解由73式,1122969611220203042001010010010701020020010NNFLFLLFEAEA解得,F1932KN26200F钢铝300解2222322BBCGALLFGALFLGALLEAEAEAEA方向向下27LFFL240,3120,360ABCDFFKNFFKNFFKN3961140101042001050010AADADFLLMMEA3966010050210010150010CGLMM396201010671010300010BELMM0670404020693GCGCCLMM解由平衡条件可求得LADABLBECC,G点的竖直位移B211AGECDF1M1M1M05M123F1F3BA211022402HFKN2/311143/5HABFFKNABFA336111103610/41110314/4ABFDM212解水压力AB杆的内力由强度条件77式故2M2M2MA3M4MB2MFABFHAF30CDB05M2MFN25252SIN30NFFFONFA260031701044802525AFKN解AB杆的轴力由强度条件77式得所以容许荷载F48KN213MAX1122NFFGAGAMAX2NFA3321163216101610043057300810201004FGALAMGH解最大轴力由强度条件77式,式中A2A2FA3M04M所以214FDCABAA216AA12L1L2F1F2A12122,FAFALLEAEA123LL1232FF解变形协调的几何关系物理关系得补充方程1252FFF1246,1111FFFF由平衡条件解得1246,1111FFAA所以31解作扭矩图如图2TTT6T2T15AAAAAT3T4T2T10NM90NM500NM/M100100200B10100（NM）AAAA3KNM2KNM4KNM1KNMC3512KNMDAA2A15KNM1KNM1525KNM32TT123D/3331242106010/3315,0,00632XPMMPAI33321047200616XPMMPAW3MAXMAX3472059108010RADG解由85式,87式和82式,MAXMAX3MIN5001630002516XPMMPAW500100300300MXNM解作扭矩图如图由37式33600NM500NM200NM600NM300NMD25D100D75D751MAX316XMD2MAX34116XMD42MAX1MAX1MAX111100100671空心圆轴增加的百分比为解原实心圆轴34从直径为300MM的实心轴中镗出一个直径为150MM的通孔而成为空心轴，问最大切应力增大了百分之几44401218013232ACABBCTALADDGOO36解由316式,把100NM,G80GPA,GGGD50MM,D35MM,L900MM代入上式,900CAAB100NM可解得A402MMLMX解MXMXX,由315式,2440DD1632LXBAPLMXTXXTLDGIGDG3712452515955119,955072,2002003010955143,955048200200TKNMTKNMTKNMTKNM3336191107920101616XMAXMDMM33122334453366119100481067,79,50201020101616DMMDDMMDMM解求外力偶矩由强度条件MXKNM11919119104838P1P1P3P4P515M175M25M15M作扭矩图如图若改用变截面1500200955955955,955573500500XABXBCPMKNMMKNMN13MAX1395510,89016XABABPMDMMDW3MAX23257310,75016BCDMMD31491955101,920180801032XABABPMDMMDGIOO3249257310,800801032BCDMMD解计算扭矩由强度条件由刚度条件取D1920MM,D280MMP2P1BP3AC310若采用同一直径,DD1920MM311600CA300T2T1B33612MAX12301,8010157161616TTDTTKND得122ACCBABACCBPPTTLTLGIGI122490603001401801032TTTRAD12523,1047TKNTKN解MXMAXT1T2,由816式,解得由强度条件ACCB解几何关系,代入几何关系，得由已知条件，TATB，得,ABACCBPAPBTLATAGIGI物理关系316LCAABTD1D2PBABPAIATTILA4211DLAD（BB757525ZYOZ3050YAO2222505030100300,38755030CCZYMM1757575/3257575/2230,7575/22575CZMM1757575/325752525/22357575/22575CYMMI1。解AB20ZY200CO20208015080120400102201501502020,46420400220150CCZYMMC2005010015050251005017591750200150100CYMM5315050917255002510ZSMM400502525050175210050250213415040025021002CYMM532100502501341115910ZSMMI2解（A）B1501501505050AZYC50ZYC150B10050505015050A442425,064644264YZYZDDDDDIII44158,2370YZICMICM3421410158391312YICM3241014237010141072558012ZICM0YZII3解AB20A号工字钢所以,ZCD/2YAOBINO20AOOO1001410014YZ4420005537,532,58,171,1851ZYICMICMBCMZCMACM22532185158171255372YZAII232ACM44200245,33,68,143ZYICMICMBCMACM268233143224522YZAII09ACMI4解A14A号槽钢,B10号工字钢AZYZAY420004817,203,70,10667ZYIICMZCMBCMACM241448172031066736851ZICM242448177020310667124662ZICM123685102956124662ZZIIA5解70708等边角钢AB所以BZYAZYAAZYYZCAA6060ZYYZCB4,012YZYZAIII40,12YZYZAIII43,096YZYZAIII430,96YZYZAIIII6证明A代入转轴公式,对于任意的角度,得B代入转轴公式,对于任意的角度,得I7DZYAZYBZYCZY解形心主轴的大致位置如图；对Z轴的惯性矩最大。20050150150502596450200150CYMM338420050501500161101212YIMMI8解A3322845020015050200505361505096425121210210ZIMMYYC150Z2005050A12121124222,222233SSFFKNMMKNM12122,2,2SSFFFMFLMFLFLFL202,33ABFKNFKN126212MMKNM解ACB412MMMM2MMMM21A212KN/M2FLL21B21FL122026,6,33SSFKNFKN2MMMM2MMMM21C218KNM6KN4MMMMFAFB75,44ABFQLFQL212732424LMMQLLQLQL,66ABQLQLFF,22ABQLQLFFDEFLL1D2122QQFAFBLL1E1QQFAFBQLLF1L12QL2QFAFB110623SQLQLQLFM1132222SQLQLQLLFQLMLQLQL36,141AAFKNMKNM02SAAAACXFXFMXFXM22492,9322SAAACBXFXFXMXFXMX解求反力列剪力方程和弯矩方程作剪力图和弯矩图如图1416972MKNMFSKN36FSKN9KN/M2M4M3KNMACBAXFAMA42LQLQQL2ABCEQLFSQLQL2/2MQL2/20ABFF202SACXLQFXQXMXX222,052SCBLXLFXQLQXLQMXQLXLQLXLE解求反力列剪力方程和弯矩方程作剪力图和弯矩图如图ALC2QL2QLBFQL2/2MQL2/2FS2QL2,0ABFQLF202SAAACXLQFXFQXMXFXX222,2SAACBLXLFXFQXMXFXQXQLF解求反力列剪力方程和弯矩方程作剪力图和弯矩图如图FAFB53,22ABFQLFQL202222SAAABXLQFXFQXMXFXX22340,42SABABBCLXLFXFQLFMXFXQLXLFXLQLL2Q2LLABQL2CG5QL/23QL/2FSQL225QL2/16M解求反力列剪力方程和弯矩方程作剪力图和弯矩图如图FAFB7,44ABFFFF202SCAXLFFFXXMXXLL2,2SAAADLXLFFLFXLFMXLXFXLLL23,22SAADBLXLFFXLFFLFLMXLXFXLFXLLFLLBF/LALCDMFL/2FL/4I3F/4F/4FSFI解求反力列剪力方程和弯矩方程作剪力图和弯矩图如图FAFB43利用剪力、弯矩与荷载集度之间的关系作下列各梁的剪力图和弯矩图。QQQL2LLFSQLAMQL2/2QL2/2FF/LFLLLLBFFFS2FL15FLFLM25KN2M2M5KN/M2M4KNMC63612M14FS611112M6,1ABFKNFKNC解求反力FAFB2FFFS11F/87F/89FL/8FL/8M3FL/2727,88ABFFFFE解求反力FLLF/LF/2LFLLEFAFB4KNM4KN/M2M2KN/M3KN/M4M2MF11/3FS613/613/324M139/3616/32713,33ABFFF解求反力FAFBQL/2QL/2IQL2/32QL2/32MQL/4QL/4QL/4FS,44ABQLQLFFI解求反力FAFBQL2/22QL3QL2/2QL2M44叠加法AQLLQLBFLLFFLLFL2FLFLFLMCFLL2FFLLLFL2FLFLMLLLQQLEQL2/2QL2/2QL2/2QL2/4M33441001800486101212ZBHIM114320100486101215810ZEIMM3MAXMAX2810148010186TCZMMPAW180502518050140825180502CYMM33224418050501801805082525180501408250857101212ZIM33MAX48108251077085710TTZMYMPAI33MAX4810147510138085710CCZMYMPAI解1矩形截面2T形截面518KNMMMM100180118050180502ZYC3460104286710YM9332001015101000301510EYMPA52梁截面最外层纤维中的正应变7104，求该梁的曲率半径。53直径D3MM的高强度钢丝，绕在直径D600MM的轮缘上，已知材料的弹性模量E200GPA，求钢丝绳横截面上的最大弯曲正应力。解M60120解2243010266710,66ZBHWM64MAX4010667102667ZMWKNM333343010205015/0152610,12212ZBHBHHWM64MAX4010526102104ZMWKNM266721041002667MMM211解A矩形截面空心矩形56A5025100252525334404055037505/027559621012ZWM4440105962102385MAXZMWKNM2440025051042106WM64401005/0551042103788MKNM37881001592385MM238537881001008412385MMM（B）总弯矩腹板承受的弯矩,腹板承受总弯矩的百分数翼缘承受弯矩的百分数4002550025B25334401803405101212ZBHIM342010015741,74140510AADZMYMPAMPAI34201001496,040510BCMPA342510015926,92640510AADZMYMPAMPAI34251001618,040510BCMPA解截面M20KNM,截面M2015325KNM,5920KNM15KNIII5M3M1M180300ADCB50ZY510117115MKNM186264188124MFSKN3M15KN/M125M186,452ABFKNFKN解求反力FAFB3MAXMAX211710813006012/6ZMMPAW3MAXMAX326410150552006012SZSZFSFMPAIBBH作剪力图和弯矩图如图60120FSMAX264KN,MMAX117KNMMAX010312046SFKN|2MAX0312011203362MKNM|3MAXMAX3033610669008/32ZMMPAW3MAXMAX24404610012233314008/4SFMPAA解12M03KN/M01KN805115121234FAFAMFF2FA/L2AFS30,2FBH33223/439248,/124212FAHHBFAHFAFALABHBHBHBHLAB0,033223/439248,/124212FAHHBFAHFAFALABHBHBHBHLAB1点M0,FSF2点MFA,FS2FA/L2A3点M0,FS2FA/L2A4点MFA,FS2FA/L2AAALFF1234H/4H/4解作剪力图和弯矩图MAX5630,05555275SSFKNFKN2280190/21601401207682280190160140CYMM32MAXMAXMAX6330102601907682/21949931026010SZFSMPAIB3963275102601001407682101779931026010SZAAZFSMPAIB解5116M5KN/M160606014050AAZYC3232642801902801909576821216014016014012076829931012ZIM151812/22706CMFF3MAX227061001503/6CZMFW13125FKN解可解得015M03M18M30KN/M12M18MFABC515517MKNM425TTCCYY33MAX82510881028876410TTMPA33MAX8410521027276410TTMPA解画弯矩图如图,故只需校核拉应力强度B截面所以此梁安全1M9KN1M1M4KNABCC截面由于B,C截面20805212020Z29/4,002/4BCFQA237/QKNMMAXMAX6/24910ZMQW167/QKNM167/QKNM解FBC9Q/4由AB梁的强度2MQ1MAB画弯矩图如图所示Q/43M/4Q/2由BC杆的强度5191212232300,0,2,AACCCCDDDDXWXLWWXLWWFLMABLCMLLDAM121200202,ABBBBBXWXLWXLWWBLAB2LFC61由弯矩图及边界条件和连续条件作桡曲线大致形状如图所示B由弯矩图及边界条件和连续条件作桡曲线大致形状如图所示解A12122323020,2,ABAAAABBBBCXLWXLWXLWWXLWWC3QL2QABL2QLLCDL3QL22QL2由弯矩图及边界条件和连续条件作桡曲线大致形状如图所示ELLLBACFL121200,2,ABBDCCCCXWXLWLXLWW由弯矩图及边界条件和连续条件作桡曲线大致形状如图所示EDFL/2FL/4DLL/2L/2BACF12121200,020,3/2,AABCCDDDDXWXLWXLWWXLWW由弯矩图及边界条件和连续条件作桡曲线大致形状如图所示2,2AAQAFQAM1AAMXFXM11AAEIWMXFXM22022AAQMXFXMXAQAXA22022AAQEIWFXMXAQAXA322422222322342AAXXQQAEIWFMXAXACXD1212,XAWWWW1212,CCDD110,0,0XW120DD120CC412CQAWEIQQA2ABACAAB,WCFAMA解B求反力分段写段弯矩方程建立微分方程积分利用连续条件和已知位移条件求积分常数62AC0XACBAX2A,32111232AAXXEIWFMCXD36BQAEI0,ABFFF10MX110EIWMX2MXFXA2EIWFXA111EIWCXD322223XAEIWFCXD1212,XAWWWW1212,CCDD10,0XW120DD22,0XAW21212FACC23A,1212CFAFAWEIEIFABAFAACCC,WCFAFB解C求反力分段写弯矩方程建立微分方程积分利用连续条件和已知位移条件求积分常数AC0XLCBAX2A5,44ABQAQAFF1AMXFX11AEIWMXFX31123AXEIWFCXD22222ABQMXFXFXAXA2222ABQEIWFXFXAXA3342222323234ABXXAQEIWFFXACXD1