電力出版社材料力學(xué)課后習(xí)題答案

解作軸力圖如下400KN340KN70KN130KNA400270340KNFF/3F/3F/3BF2F/3F/3213FF2FC3F2FLL/2LFFF/LFDF2FFN題71圖2F2FFFF2FFE2FF/LF/L23A2M2M4M12MABC1MDQ10KN/M12FAFBFCXFCYFAAFN2CDQ10KN/M1FN2FN1DN10404104236422ICMFKN120,5/2407,IXFFFKN3111636410/354,115010FAMPA3222640710/317115010FAMPA解求反力,FAFB40KN由結(jié)點D的平衡取一半如圖分析23B1M1M1M1MABECDQ3KN/M123AF1ECDQ3KN/MF2F1CF310,3315/1135AMFKN310,2191IXFFFKN230,1/2135IYFFFKN31161326336135101598501013510225600101911038250010FMPAAMPAMPA解去掉1桿的約束如圖,用約束力F1代之,由由結(jié)點C的平衡3MAX6MIN14103502010410FMPAA368102673010BCMPA解AFNF10KN,B軸力圖如圖,AB17KN9KN16KNCD11B22B0FFA24BKN81724（B36241020012010CDMPA3617102138010ABMPA361210403010BCMPA3618104540010ABMPAMAX45BCMPA（C）,12KN50CMABCC50CM解由73式,1122969611220203042001010010010701020020010NNFLFLLFEAEA解得,F1932KN26200F鋼鋁300解2222322BBCGALLFGALFLGALLEAEAEAEA方向向下27LFFL240,3120,360ABCDFFKNFFKNFFKN3961140101042001050010AADADFLLMMEA3966010050210010150010CGLMM396201010671010300010BELMM0670404020693GCGCCLMM解由平衡條件可求得LADABLBECC,G點的豎直位移B211AGECDF1M1M1M05M123F1F3BA211022402HFKN2/311143/5HABFFKNABFA336111103610/41110314/4ABFDM212解水壓力AB桿的內(nèi)力由強(qiáng)度條件77式故2M2M2MA3M4MB2MFABFHAF30CDB05M2MFN25252SIN30NFFFONFA260031701044802525AFKN解AB桿的軸力由強(qiáng)度條件77式得所以容許荷載F48KN213MAX1122NFFGAGAMAX2NFA3321163216101610043057300810201004FGALAMGH解最大軸力由強(qiáng)度條件77式,式中A2A2FA3M04M所以214FDCABAA216AA12L1L2F1F2A12122,FAFALLEAEA123LL1232FF解變形協(xié)調(diào)的幾何關(guān)系物理關(guān)系得補(bǔ)充方程1252FFF1246,1111FFFF由平衡條件解得1246,1111FFAA所以31解作扭矩圖如圖2TTT6T2T15AAAAAT3T4T2T10NM90NM500NM/M100100200B10100（NM）AAAA3KNM2KNM4KNM1KNMC3512KNMDAA2A15KNM1KNM1525KNM32TT123D/3331242106010/3315,0,00632XPMMPAI33321047200616XPMMPAW3MAXMAX3472059108010RADG解由85式,87式和82式,MAXMAX3MIN5001630002516XPMMPAW500100300300MXNM解作扭矩圖如圖由37式33600NM500NM200NM600NM300NMD25D100D75D751MAX316XMD2MAX34116XMD42MAX1MAX1MAX111100100671空心圓軸增加的百分比為解原實心圓軸34從直徑為300MM的實心軸中鏜出一個直徑為150MM的通孔而成為空心軸，問最大切應(yīng)力增大了百分之幾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解求外力偶矩由強(qiáng)度條件MXKNM11919119104838P1P1P3P4P515M175M25M15M作扭矩圖如圖若改用變截面1500200955955955,955573500500XABXBCPMKNMMKNMN13MAX1395510,89016XABABPMDMMDW3MAX23257310,75016BCDMMD31491955101,920180801032XABABPMDMMDGIOO3249257310,800801032BCDMMD解計算扭矩由強(qiáng)度條件由剛度條件取D1920MM,D280MMP2P1BP3AC310若采用同一直徑,DD1920MM311600CA300T2T1B33612MAX12301,8010157161616TTDTTKND得122ACCBABACCBPPTTLTLGIGI122490603001401801032TTTRAD12523,1047TKNTKN解MXMAXT1T2,由816式,解得由強(qiáng)度條件ACCB解幾何關(guān)系,代入幾何關(guān)系，得由已知條件，TATB，得,ABACCBPAPBTLATAGIGI物理關(guān)系316LCAABTD1D2PBABPAIATTILA4211DLAD（BB757525ZYOZ3050YAO2222505030100300,38755030CCZYMM1757575/3257575/2230,7575/22575CZMM1757575/325752525/22357575/22575CYMMI1。解AB20ZY200CO20208015080120400102201501502020,46420400220150CCZYMMC2005010015050251005017591750200150100CYMM5315050917255002510ZSMM400502525050175210050250213415040025021002CYMM532100502501341115910ZSMMI2解（A）B1501501505050AZYC50ZYC150B10050505015050A442425,064644264YZYZDDDDDIII44158,2370YZICMICM3421410158391312YICM3241014237010141072558012ZICM0YZII3解AB20A號工字鋼所以,ZCD/2YAOBINO20AOOO1001410014YZ4420005537,532,58,171,1851ZYICMICMBCMZCMACM22532185158171255372YZAII232ACM44200245,33,68,143ZYICMICMBCMACM268233143224522YZAII09ACMI4解A14A號槽鋼,B10號工字鋼AZYZAY420004817,203,70,10667ZYIICMZCMBCMACM241448172031066736851ZICM242448177020310667124662ZICM123685102956124662ZZIIA5解70708等邊角鋼AB所以BZYAZYAAZYYZCAA6060ZYYZCB4,012YZYZAIII40,12YZYZAIII43,096YZYZAIII430,96YZYZAIIII6證明A代入轉(zhuǎn)軸公式,對于任意的角度,得B代入轉(zhuǎn)軸公式,對于任意的角度,得I7DZYAZYBZYCZY解形心主軸的大致位置如圖；對Z軸的慣性矩最大。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彎矩方程作剪力圖和彎矩圖如圖1416972MKNMFSKN36FSKN9KN/M2M4M3KNMACBAXFAMA42LQLQQL2ABCEQLFSQLQL2/2MQL2/20ABFF202SACXLQFXQXMXX222,052SCBLXLFXQLQXLQMXQLXLQLXLE解求反力列剪力方程和彎矩方程作剪力圖和彎矩圖如圖ALC2QL2QLBFQL2/2MQL2/2FS2QL2,0ABFQLF202SAAACXLQFXFQXMXFXX222,2SAACBLXLFXFQXMXFXQXQLF解求反力列剪力方程和彎矩方程作剪力圖和彎矩圖如圖FAFB53,22ABFQLFQL202222SAAABXLQFXFQXMXFXX22340,42SABABBCLXLFXFQLFMXFXQLXLFXLQLL2Q2LLABQL2CG5QL/23QL/2FSQL225QL2/16M解求反力列剪力方程和彎矩方程作剪力圖和彎矩圖如圖FAFB7,44ABFFFF202SCAXLFFFXXMXXLL2,2SAAADLXLFFLFXLFMXLXFXLLL23,22SAADBLXLFFXLFFLFLMXLXFXLFXLLFLLBF/LALCDMFL/2FL/4I3F/4F/4FSFI解求反力列剪力方程和彎矩方程作剪力圖和彎矩圖如圖FAFB43利用剪力、彎矩與荷載集度之間的關(guān)系作下列各梁的剪力圖和彎矩圖。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疊加法AQLLQLBFLLFFLLFL2FLFLFLMCFLL2FFLLLFL2FLFLMLLLQQLEQL2/2QL2/2QL2/2QL2/4M33441001800486101212ZBHIM114320100486101215810ZEIMM3MAXMAX2810148010186TCZMMPAW180502518050140825180502CYMM33224418050501801805082525180501408250857101212ZIM33MAX48108251077085710TTZMYMPAI33MAX4810147510138085710CCZMYMPAI解1矩形截面2T形截面518KNMMMM100180118050180502ZYC3460104286710YM9332001015101000301510EYMPA52梁截面最外層纖維中的正應(yīng)變7104，求該梁的曲率半徑。53直徑D3MM的高強(qiáng)度鋼絲，繞在直徑D600MM的輪緣上，已知材料的彈性模量E200GPA，求鋼絲繩橫截面上的最大彎曲正應(yīng)力。解M60120解2243010266710,66ZBHWM64MAX4010667102667ZMWKNM333343010205015/0152610,12212ZBHBHHWM64MAX4010526102104ZMWKNM266721041002667MMM211解A矩形截面空心矩形56A5025100252525334404055037505/027559621012ZWM4440105962102385MAXZMWKNM2440025051042106WM64401005/0551042103788MKNM37881001592385MM238537881001008412385MMM（B）總彎矩腹板承受的彎矩,腹板承受總彎矩的百分?jǐn)?shù)翼緣承受彎矩的百分?jǐn)?shù)4002550025B25334401803405101212ZBHIM342010015741,74140510AADZMYMPAMPAI34201001496,040510BCMPA342510015926,92640510AADZMYMPAMPAI34251001618,040510BCMPA解截面M20KNM,截面M2015325KNM,5920KNM15KNIII5M3M1M180300ADCB50ZY510117115MKNM186264188124MFSKN3M15KN/M125M186,452ABFKNFKN解求反力FAFB3MAXMAX211710813006012/6ZMMPAW3MAXMAX326410150552006012SZSZFSFMPAIBBH作剪力圖和彎矩圖如圖60120FSMAX264KN,MMAX117KNMMAX010312046SFKN|2MAX0312011203362MKNM|3MAXMAX3033610669008/32ZMMPAW3MAXMAX24404610012233314008/4SFMPAA解12M03KN/M01KN805115121234FAFAMFF2FA/L2AFS30,2FBH33223/439248,/124212FAHHBFAHFAFALABHBHBHBHLAB0,033223/439248,/124212FAHHBFAHFAFALABHBHBHBHLAB1點M0,FSF2點MFA,FS2FA/L2A3點M0,FS2FA/L2A4點MFA,FS2FA/L2AAALFF1234H/4H/4解作剪力圖和彎矩圖MAX5630,05555275SSFKNFKN2280190/21601401207682280190160140CYMM32MAXMAXMAX6330102601907682/21949931026010SZFSMPAIB3963275102601001407682101779931026010SZAAZFSMPAIB解5116M5KN/M160606014050AAZYC3232642801902801909576821216014016014012076829931012ZIM151812/22706CMFF3MAX227061001503/6CZMFW13125FKN解可解得015M03M18M30KN/M12M18MFABC515517MKNM425TTCCYY33MAX82510881028876410TTMPA33MAX8410521027276410TTMPA解畫彎矩圖如圖,故只需校核拉應(yīng)力強(qiáng)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梁的強(qiáng)度2MQ1MAB畫彎矩圖如圖所示Q/43M/4Q/2由BC桿的強(qiáng)度5191212232300,0,2,AACCCCDDDDXWXLWWXLWWFLMABLCMLLDAM121200202,ABBBBBXWXLWXLWWBLAB2LFC61由彎矩圖及邊界條件和連續(xù)條件作橈曲線大致形狀如圖所示B由彎矩圖及邊界條件和連續(xù)條件作橈曲線大致形狀如圖所示解A12122323020,2,ABAAAABBBBCXLWXLWXLWWXLWWC3QL2QABL2QLLCDL3QL22QL2由彎矩圖及邊界條件和連續(xù)條件作橈曲線大致形狀如圖所示ELLLBACFL121200,2,ABBDCCCCXWXLWLXLWW由彎矩圖及邊界條件和連續(xù)條件作橈曲線大致形狀如圖所示EDFL/2FL/4DLL/2L/2BACF12121200,020,3/2,AABCCDDDDXWXLWXLWWXLWW由彎矩圖及邊界條件和連續(xù)條件作橈曲線大致形狀如圖所示2,2AAQAFQAM1AAMXFXM11AAEIWMXFXM22022AAQMXFXMXAQAXA22022AAQEIWFXMXAQAXA322422222322342AAXXQQAEIWFMXAXACXD1212,XAWWWW1212,CCDD110,0,0XW120DD120CC412CQAWEIQQA2ABACAAB,WCFAMA解B求反力分段寫段彎矩方程建立微分方程積分利用連續(xù)條件和已知位移條件求積分常數(shù)62AC0XACBAX2A,32111232AAXXEIWFMCXD36BQAEI0,ABFFF10MX110EIWMX2MXFXA2EIWFXA111EIWCXD322223XAEIWFCXD1212,XAWWWW1212,CCDD10,0XW120DD22,0XAW21212FACC23A,1212CFAFAWEIEIFABAFAACCC,WCFAFB解C求反力分段寫彎矩方程建立微分方程積分利用連續(xù)條件和已知位移條件求積分常數(shù)AC0XLCBAX2A5,44ABQAQAFF1AMXFX11AEIWMXFX31123AXEIWFCXD22222ABQMXFXFXAXA2222ABQEIWFXFXAXA3342222323234ABXXAQEIWFFXACXD1