中國石油大學工程力學答案合集PPT課件

1、2021-10-27Statics-Problems-1靜力學第1章作業(yè)/P23-1.1(g)*:第1頁/共96頁2021-10-27Statics-Problems-2靜力學第1章作業(yè)/P24-1.2(k):第2頁/共96頁2021-10-27Statics-Problems-3靜力學第1章作業(yè)/P27-1.3(c):第3頁/共96頁2021-10-27Statics-Problems-4靜力學第1章作業(yè)/P27-1.3(e)*:第4頁/共96頁2021-10-27Statics-Problems-5靜力學第2章作業(yè)/P58-2.3*:)(903060cos40coskNPRxSolutio

2、n:由此得到如下方程組：)( 060sin40sinkNPRy320sin100cosPP解得：)( 8 .105740kNP)( 1 .1953tan1第5頁/共96頁2021-10-27Statics-Problems-6靜力學第2章作業(yè)/P60-2.14*:Solution:按定義有:2211)(rFFrrFFrFMOSo, expressions (a), (d), (e) are valid.第6頁/共96頁2021-10-27Statics-Problems-7靜力學第2章作業(yè)/P61-2.20:Solution:(a)ar)(16. 04 . 048. 0mkjira)(6812

3、NkjiFjaIn vector form:)(96. 0)(mNjFMa)(96. 0010681216. 04 . 048. 0)(mNFrFMaaa第7頁/共96頁2021-10-27Statics-Problems-8)( 0100681216. 04 . 06 . 0)(mNFrFMzzzzr(b)(16. 04 . 06 . 0mkjirz)(6812NkjiFkzIn vector form:)( 0)(mNFMz第8頁/共96頁2021-10-27Statics-Problems-9+0 xxFRkNFRyy16161220mkNMM.36716485122200LRRLMy0

4、由)(25. 216360mRMLxySolution:靜力學第2章作業(yè)/P63-2.28*:第9頁/共96頁2021-10-27Statics-Problems-10與水平面夾角為，與水平面夾角為21TT1585 . 18 . 0tan178sin1715cos432 . 19 . 0tan53sin54cos0coscos21TTFX+0505 .142sinsin21TTFy)(1701NT )( 5 .1872NT （a)Solution:靜力學第2章作業(yè)/P63-2.30*:第10頁/共96頁2021-10-27Statics-Problems-11(b)+ARMC) 8 . 05

5、. 09 . 0(cos) 2 . 12 . 15 . 1 (sin22TT)2 . 15 . 1 (505 . 15 .142).( 0mN第11頁/共96頁2021-10-27Statics-Problems-12OOriginal forces and couple can be reduced to a force and a couple acting at point O:靜力學第2章作業(yè)/P64-2.32:RSolution:RC)(31030cos20NRx)(506030sin20kNRy)( 3 .23353212130sin205 . 030cos205 . 060mNC

6、R于是得到合力的大小、方向和作用線位置（如圖所示）：)( 9 .5272022NRRRyx)( 1 .1953tantan11yxRR)(441. 0mRCdR)(467. 0mRCdyRx)(348. 1mRCdxRyyddxd(或,或)yRRxR第12頁/共96頁2021-10-27Statics-Problems-13+0MCR0 xF0yF0zFNPx80NPy60NPz90kjiP906080mNM.392 . 0603 . 0900Solution:靜力學第2章作業(yè)/P64-2.34*:第13頁/共96頁2021-10-27Statics-Problems-14靜力學第3章作業(yè)/P

7、85-3.2*:Solution:RAyRByFrom the FBD of the beam in Fig:0BM+023129212AyR)( 0 kNRAyRBx3kN2m0 xF+)( 0 kNRBx0yF+032ByAyRR)( 1 kNRBy第14頁/共96頁2021-10-27Statics-Problems-15靜力學第3章作業(yè)/P86-3.11:Solution:GPGLStudying person and ladder, draw FBD(a) x=1.5m0AM+022BLPRGxGRBRAyRAx744.8(N)762gGxGRLPB0 xF+053AxBRR446.

8、88(N)53BAxRR第15頁/共96頁2021-10-27Statics-Problems-160yF+054AyBLPRRGGGPGLRBRAyRAx344.96(N)517651521254gGGxGxGGGRLPLPLPAy(b) 051521LPAyGGxR)( 6 . 2215 . 2mGGxPL第16頁/共96頁2021-10-27Statics-Problems-17靜力學第3章作業(yè)/P87-3.18:Solution:300N0.5mRDRE(1) Studying entire structureFrom the FBD of entire structure in Fi

9、g:0DM+01 . 13006 . 1ER206.25(N)ER(2) Studying member ABFrom the FBD of member AB in Fig:RBxRByRAxRAy300N0.5m1.6m0BM+05 . 03006 . 1AyR93.75(N)AyR第17頁/共96頁2021-10-27Statics-Problems-18(N)00434AyEAxRRR(3) Studying member ACEFrom the FBD of member ACE in Fig:0.8mRERAxRAyAECRCxRCy0.8m0.6m0.6m0CM+06 . 08

10、. 08 . 0AxAyERRR0 xF+0CxAxRR)(400 NRRAxCx0yF+0ECyAyRRR-112.5(N)EAyECyRRRR.8(N)10422AyAxARRR.5(N)15422CyCxCRRR第18頁/共96頁2021-10-27Statics-Problems-19靜力學第3章作業(yè)/P90-3.28(f)*:Solution:(1) Studying member CDFrom the FBD of member CD in Fig:顯然，本題中在約束A、B、C、D處沒有水平方向的作用力RCRMRCR2qC3m1m0CM+0412DRMq42MqRD0DM+0324

11、MqRC423MqRC第19頁/共96頁2021-10-27Statics-Problems-20(2) Studying member ABCRARCRB2qAC1m2mRARCRB1mFrom the FBD of member CD in Fig:0AM+04322CBRqR26MqRB0BM+02122CARqR425MqRA第20頁/共96頁2021-10-27Statics-Problems-21靜力學第3章作業(yè)/P91-3.40:Solution:(1) From the FBD of joint A in Fig:APFABFAD1250 xF+052ADABFF0yF+051

12、ADFP)(5TensionPFAD)(2nCompressioPFAB第21頁/共96頁2021-10-27Statics-Problems-22(2) From the FBD of joint B in Fig:BPFBCFBDFAB0 xF+0BCABFF0yF+0BDFP)(2nCompressioPFBC)(TensionPFBD第22頁/共96頁2021-10-27Statics-Problems-23)(25nCompressioPFCD(3) From the FBD of joint D in Fig:DFBD0 xF+0525252DECDADFFF0yF+0515151

13、DECDADBDFFFF)(253TensionPFDEFDEFADFCD125125125第23頁/共96頁2021-10-27Statics-Problems-24(4) From the FBD of joint C in Fig:CRCFCEFBCFCD1250yF+051CECDFF)(21TensionPFCE第24頁/共96頁2021-10-27Statics-Problems-25)(413)(413nCompressioPFTensionPFGHCH靜力學第3章作業(yè)/P92-3.43*:Solution:(1) From the FBD of joint D in Fig:D

14、PFCDFDH0yF+0DHFPPFDH(1) From the FBD of joint in Fig:HFDHFGFC231323130 xF+0133133GHCHFF0yF+0132132DHGHCHFFF第25頁/共96頁2021-10-27Statics-Problems-26Solution:lbF54952 . 1合+060sin6 . 3290CDxFFM合04125BzAzyFFGM04125ByAyzFFFM合060sin0CDBzAzzFGFFF060cos0CDByAyyFFFFF合+)(33027lbFAy)(321527lbFBy)( 5 .612123lbFBz

15、)(51lbFAz)(340lbFCD靜力學第4章作業(yè)/P111-4.2*:第26頁/共96頁2021-10-27Statics-Problems-27靜力學第4章作業(yè)/P113-4.12*:Solution:RAzRAyRByRBxRCxRCzFrom the FBD of bent pipe and cable in Fig:0 xM04 . 08 . 0CzByAyRRR0yM08 . 04 . 08 . 0PRRBxAz0zM08 . 0CxAyRR0 xF0PRRCxBx0yF0PRRByAy0zF0CzAzRR第27頁/共96頁2021-10-27Statics-Problems-

16、2808 . 00011100008 . 00004 . 08 . 00014 . 0008 . 0010010001001100100PRRRRRRCxCzByBxAzAy無解??？第28頁/共96頁2021-10-27Statics-Problems-290 xM0yM0zM04030CBTT0300210DzCFT030090DyBFT0yF0zF0BDyAyTFF0CDzAzTFF)(60 NTC)(56 NFAy)(18 NFAz)(24 NFDy)(42 NFDzSolution:靜力學第4章作業(yè)/P113-4.13*:第29頁/共96頁2021-10-27Statics-Probl

17、ems-30靜力學第4章作業(yè)/P114-4.14(b):Solution:S令形心坐標為(xc, yc), 按定義有:SScdSxdSxSScdSydSy)(51624023410402341ftdxxdydxdSxS)(76434023410402341ftdxxxdyxdxxdSxS)(23234030402341ftdxxydydxydSxS)(720ftdSxdSxSSc)(85ftdSydSySSc第30頁/共96頁2021-10-27Statics-Problems-31靜力學第4章作業(yè)/P115-4.24:Solution:已知crucible的密度rc=2000kg/m3, 外

18、徑R=0.28m, 內徑r=0.2m; iron的密度ri=7200kg/m3OStudying crucible，Drawing FBD in Fig.WiWcGiGcFyFx0ABM+0sincosccOGWOCFtanOCOGWFccgVWcccrROCcccVrRVrrRROG44334132833283第31頁/共96頁2021-10-27Statics-Problems-32)(tan250tan4tan141tan4444NRrRgRVrRgVOCOGWFccccccrrgVWcccrROCcccVrRVrrRROG44334132833283OWiWcGiGcFyFx第32頁/

19、共96頁2021-10-27Statics-Problems-33Solution:端圓心為原點：軸如課本所示，木柄底、軸，以木柄軸線為yxz)/(1033. 8)225(3040304521304090/(919. 0362mmkgvmsssteelr)(8167. 0)904030(1kgmsteelr)(1227. 0)225(30(22kgmsteelr)(2250. 0)454030(3kgmsteelr)(151mmy)(02mmy)(453mmy )(1953mmz )(1952mmz )(1951mmz x=0 )(088. 2432144332211mmmmmmymymymy

20、my)(099.04kgm)( 04mmy )(1054mmz )(25.186432144332211mmmmmmzmzmzmzmz靜力學第4章作業(yè)/P116-4.27*:第33頁/共96頁2021-10-27Statics-Problems-34xyO2xyO1靜力學第5章作業(yè)/P131-5.2:解：3075. 0s75. 0s36lb20lb18in假設系統(tǒng)是靜力平衡的先研究圓柱，有030sin00011111fGPXrfMOP1G1G2N1N2f2f1P1得1121GP 再研究方塊，有030cos0030sin022221NGYfGPX得)(28)(21212lbGGf而)(4 .23

21、232222max2lbGNfmax22ff 因，所以系統(tǒng)不能保持靜力平衡第34頁/共96頁2021-10-27Statics-Problems-35Discussion:1. 方塊會不會先翻倒?3075. 0s75. 0s36lb20lb18in2. 如果圓柱和方塊的接觸面不光滑,會怎樣?xyO2G2N2f2P1x2x2=?第35頁/共96頁2021-10-27Statics-Problems-36Solution:總體分析: 0CAff分析木條 :0 xF0yF0AM0SinNCosffBBA02GSinfCosNNBBA025 . 15 . 012GCosNB圓盤分析 :0OM03 .

22、03 . 0CBff解之得： ASAANfBSBBNfCSCCNfNNA64.39NNB27.150NNC27.390NfffCBA.22.0655. 0AS15. 0BS056. 0CS0AM025 . 15 . 01121CosGGNC0CM01)25 . 15 . 01 (2ANCosG0 xF達到最大靜摩擦力時 :解之得： 靜力學第5章作業(yè)/P132-5.10*:第36頁/共96頁2021-10-27Statics-Problems-379000N/mBCDBCD靜力學第5章作業(yè)/P132-5.12:解：P假設系統(tǒng)是靜力平衡的先研究樁子CBD，有NBNBNAfAfBRxRy)(9000

23、05 . 1) 331()9000321(0NNNMBBC再研究桿AB，有000008 . 08 . 20PffYNNXPfMBABABA)(450075)(225072maxmaxNNfPfNNfPfAAsAABBsBB5 . 0s25. 0sP9000N/m1.5m2m3m0.8mABCDABfBAB第37頁/共96頁2021-10-27Statics-Problems-38只有當fAfAmax或fBfBmax時，桿AB才能移動，由以上計算結果可知，當P6300(N)，桿AB在A點處開始移動第38頁/共96頁2021-10-27Statics-Problems-39靜力學第5章作業(yè)/P13

24、3-5.18*:Solution:0601CosPN0601SinPmgN0 xF0yFNP8 .550105 . 0311601322mgSinPmg0OMNP9 .1032NP9 .103因此 :第39頁/共96頁2021-10-27Statics-Problems-40Solution:020GSinfP020GCosNNf 05 . 0208 . 020PhGCosGSinNP3 .374木塊翻轉臨界時：滑動時：mh192. 1靜力學第5章作業(yè)/P133-5.19*:第40頁/共96頁2021-10-27Statics-Problems-41BANN BAff AANf當P=600 l

25、b時： 0yF06005 . 7cos5 . 7cos5 . 7sin5 . 7sin0000BABAffNNBBNf當P=200 lb時： 0yF02005 . 7cos5 . 7cos5 . 7sin5 . 7sin0000BABAffNN2633. 0Solution:靜力學第5章作業(yè)/P134-5.22*:第41頁/共96頁2021-10-27MOM-Problems-42第六次作業(yè)/P34-1.1(b):SolutionPPPP332211PPPPPPPPPPPPN1N2N3N1=PN2=0N3=PNPP001PNFx001PPNFx001PPPNFx第42頁/共96頁2021-10

26、-27MOM-Problems-43第六次作業(yè)/P34-1.6:SolutionPpd1DPpp6Pb研究活塞桿，有)(4212dDpP 12114dP )( 1 .181211MPadDp研究右擋蓋，有)(46212dDpPb 2224dPb )(50. 6621222MPadDdp得)(50. 6maxMPap第43頁/共96頁2021-10-27MOM-Problems-44第六次作業(yè)/P34-1.7*:SolutionPBPsPwBSteelWoodP30AC研究節(jié)點B，有030sin0030cos0PPYPPXswsPPPPws3,2研究桿BC，有 )(0 .48212222222k

27、NAPAPAPs研究桿AB，有 )(4 .40313111111kNAPAPAPw取)(4 .40kNP 第44頁/共96頁2021-10-27MOM-Problems-45第六次作業(yè)/P35-1.13*:SolutionBPPCASteelAluminumL1L21m)(4 . 0)(405. 01021075. 010110222931mmmAEPLs根據(jù)題意列出方程組：解得)(25. 0)(75. 021mEEELmEEELassassAEPLAEPLLLas21211總伸長為第45頁/共96頁2021-10-27MOM-Problems-46dxx第六次作業(yè)/P37-1.14*:Sol

28、utionLBCAc(1) A formula for the downward displacement C of point CxPLxLWPPEAPdxcL0cdxEALxLWcL0dxEALWxLcEALcLW222(2) The elongation B of the entire barBEALLW2022EAWL2第46頁/共96頁2021-10-27MOM-Problems-47(3) The ratio r of the elongation of the upper half of the bar to the elongation of the lower half of

29、 the barThe elongation of the upper half of the barufEALLLW2222EAWL83lfEAWLufB8The elongation of the lower half of the barSo3lfufr第47頁/共96頁2021-10-27MOM-Problems-48第六次作業(yè)/P37-1.21:SolutionE1E2eebbPPP1P1P2P2(1) 靜力平衡方程：研究右擋板，有PebPbPMPPPXo22002121(2) 變形協(xié)調方程：PeP1P2O2211221121EPEPAELPAELPLL(3) 求解方程組：21212

30、12121212221112EEAPAPEEEEbeEEPEPEEPEP第48頁/共96頁2021-10-27MOM-Problems-49abLAyRP1N2NAxR第六次作業(yè)/P37-1.23*:SolutionPhh2B1A2CDabL(1) Equation of equilibrium BACDPLbNaN21(2) Equation of compatibility ba21(3) Force-displacement relations EAhN11EAhN222得補充方程baNN212第49頁/共96頁2021-10-27MOM-Problems-50PLbNaN21baNN2

31、12解得有22122baPLaN2222baPLbN22222112222baEAPhLbaPLaEAhaLEAhNaLaLB第50頁/共96頁2021-10-27MOM-Problems-51第六次作業(yè)/P37-1.24*:Solution2a2a 變形過程中3根桿的長度始終滿足下面的關系：223212241)LL(21La)LL(31L)L2LL2L(21L2L312331122而3根桿的長度變化滿足下面的關系：(1) 靜力平衡方程：研究節(jié)點A，有132P3030APAP1P3P2030sin30sin0030cos30cos031321PPPYPPPX(2) 變形協(xié)調方程：第51頁/共9

32、6頁2021-10-27MOM-Problems-5202320323213311332211PAAAAA(3) 求解方程組：)(30cos31)(31331122333111222APAPAPEALPEALPEALPMPaPAAAAAAAAMPaPAAAAAAAAMPaPAAAAAAAA6 .8633224338 .26332232328 .1263322334133221213133221132133221321030sin30sin0030cos30cos031321PPPYPPPX第52頁/共96頁2021-10-27MOM-Problems-53F F30ABC30D123xyP P

33、A1P2P3PxyAAxy將將A A點的位移分量向各桿點的位移分量向各桿投影投影. .得得cossin1xylxl2cossin3xylcos2213lll變形關系為變形關系為 2133 lll(4) 討論：3根桿長度變化關系的另一種求法：第53頁/共96頁2021-10-27MOM-Problems-54RBRA第六次作業(yè)/P37-1.26:SolutionA1A2aa(1) Equation of equilibrium 0BARR(2) Equation of compatibility A1A2RB0TTA1A2TT0RTA1A2RBR(3) Force-displacement re

34、lations TaTaTaT2212111AAEaREAaREAaRBBBR得補充方程011221AAEaRTaB第54頁/共96頁2021-10-27MOM-Problems-55解得21112AATERBMPaAATEARB667. 010511020020105 .12212982111MPaAATEARB333. 051011020020105 .12212981222第55頁/共96頁2021-10-27MOM-Problems-56T2T1材料力學第2章作業(yè)/P58-2.1(b) :Solutionm3mmm3mmmm23T2T2mmm1T001mTMx0302mmTMx第56頁

35、/共96頁2021-10-27MOM-Problems-57材料力學第2章作業(yè)/P59-2.11:Solution123ABC400500N1N2N3(1) 123ABCm1m2m3)(214. 45003007024N7024m)(810. 25002007024N7024m)(024. 75005007024N7024m332211mkNnmkNnmkNn)(214. 4T)(024. 7T3BC1ABmkNmmkNm對于AB段， )(0 .801616/33maxmmTddTWTABABABABtABAB由強度條件得 )(6 .843218032/1801804242maxmmGTddG

36、TGITABABABABPABAB由剛度條件得 第57頁/共96頁2021-10-27MOM-Problems-58)(85 mmdAB取 )(4 .671616/33maxmmTddTWTBCBCBCBCtBCBC由強度條件得 )(5 .743218032/1801804242maxmmGTddGTGITCBBCBCBCPBCBC由剛度條件得 對于BC段，)(75 mmdAB取(2) )(85 mmdAB取(3) 齒輪1安放在齒輪2、3中間。123ABC400500N1N2N3123ABCm1m2m3第58頁/共96頁2021-10-27MOM-Problems-59材料力學第2章作業(yè)/P5

37、9-2.14:Solution方法一（傳統(tǒng)法）：L/4L/4L/2AB3T0T01. 解除某些約束，施加相應約束反力，使之成為靜定基；2. 扭矩與變形協(xié)調方程：04/32/00PPPBGILTGILTGILT得TB=5T0/43. 外扭矩3T0和T0作用處的轉角分別為：PPPPPBPPPPBGILTGILTGILTGILTGILTGILTGILTGILTGILT1611854/32/2116114/3212/4101max000200013T03T0T0TBT0TB=第59頁/共96頁2021-10-27MOM-Problems-60材料力學第2章作業(yè)/P59-2.14:Solution方法二

38、（船長法）：L/4L/4L/2AB3T0T03T0T0TATB1. 解除約束，施加約束反力；2. 扭矩與變形協(xié)調方程：右起第一段：TB右起第二段：TB- T0右起第二段：TB- 4T004/)4(4/)(2/00PBPBPBGILTTGILTTGILT得TB=5T0/43. 可見，右起第一段和第二段的扭矩都大于0，最大轉角發(fā)生在外扭矩3T0作用處：PPBPBGILTGILTTGILT16114/)(2/00max第60頁/共96頁2021-10-27MOM-Problems-61材料力學第3章作業(yè)/P72-3.1(c):BqaqACaaqa21122Solutionqaqqa2V1M12112

39、321,2qaaqaaqaMqaqaqaV對于1-1截面，有qaqqa2V2M222222121,2qaaqaaqaqaMqaqaqaV對于2-2截面，有第61頁/共96頁2021-10-27MOM-Problems-62qa補充：內力圖BqaqACaaqa2Vqa2x-221qa223qa225qax第62頁/共96頁2021-10-27MOM-Problems-63(1) Shear-Force and Bending-Moment Equations 材料力學第3章作業(yè)/P73-3.2(a):Solution2PBACaaPa2PBACPaxxaaVM020PVYPV20)(20Paxa

40、PMMOPaPxM2ax02PBACPaxxa2VMax000VY0V00PaMMOPaM axa2axa2第63頁/共96頁2021-10-27MOM-Problems-64(2) Shear-Force and Bending-Moment Diagrams PV2PaPxM2ax00VPaM axa22PBACaaPaxxVP2-PaPa(3) Maximum Shear Forces and Bending MomentsPV2maxPaMmax第64頁/共96頁2021-10-27MOM-Problems-65材料力學第3章作業(yè)/P73-3.3(b)*:qABCL/2L/2Solut

41、ionxVMVM) 2/0(,31312/21,2/2132LxxLqxLqxxMxLqLqxxV,2424)2(21)2()232(221,4)2(22122xqxqLqLLxLxqLxqLMqxqLqLxqLV)2/(LxLxxM242qL2472qL控制截面法？V4qL43qL第65頁/共96頁2021-10-27MOM-Problems-66Solution材料力學第4章作業(yè)/P90-4.6*:PPACDB2m2m2mNo20a1. 求約束反力2. 求彎矩作出彎矩圖RARB0006420BAABRRPPYRPPMPRPRBA313103223220BBDACAMPRMPRMM3. 求許

42、用載荷MP32P32 )(88.5610160102373266maxmaxkNPPWM由強度條件得危險截面在C或D，有,32maxPM查表得),(2373cmW 第66頁/共96頁2021-10-27MOM-Problems-67材料力學第4章作業(yè)/P91-4.8:SolutionA T-beam made of cast iron in pure bending. Knowing the ratio of tensile allowable stress to compressive allowable stress is t/c=1/4. Determine the reasonable

43、 width b of the flange. bzC3060400MMz3403060230340303060bbycy)(1703910030mmbb tzctcIMy310 czcccIyM310400 4400tcccyymmyc80801703910030bbmmb510第67頁/共96頁2021-10-27MOM-Problems-68材料力學第4章作業(yè)/P91-4.13:Solution1. 求剪力和彎矩PMMPVABAB0作出剪力圖和彎矩圖PV2. 根據(jù)膠合面強度條件求P1mPAB505050100G整個截面都是危險截面，考慮剪應力互等定理，有 )(825. 31034. 00

44、1125. 0)025. 0415. 0(1215. 01 . 026223kNPPPgGMP第68頁/共96頁2021-10-27MOM-Problems-69PV1mPAB505050100GMP3. 根據(jù)截面剪應力強度條件求P整個截面都是危險截面，有 )(1010101. 015. 01 . 0236maxkNPPP4. 根據(jù)截面正應力強度條件求PA截面是危險截面，有 )(75. 31010000375. 0615. 01 . 062maxkNPPP5. 綜上所述，取)(75. 3kNP 第69頁/共96頁2021-10-27MOM-Problems-70 xLq0q0LBAyx材料力學

45、第5章作業(yè)/P110-5.3(c)*:Solution1. 求約束反力03221003121000LRLLqMLRLLqMBAABRARBLqRLqRBA0031612. 求彎矩方程OARAV Mx)1(613121)(300 xLLxqxxxLqxRxMA第70頁/共96頁2021-10-27MOM-Problems-713. 求撓曲線方程：2150301402030012036241266)(CxCxLqxLqEIvCxLqxLqvEIxLqxLqxMvEI 代入邊界條件：000vLxvx時，時，得：360703012LqCC)3107(3604230 xLLxLEIxqv最大撓度：兩端轉

46、角：LLx51933. 015810EILqLv7685)21(40中間撓度：)15307(3604230 xLLxLEIqvEILqA360730EILqB4530EILqxvv76801. 5)(400max0)(0 xv取合理解第71頁/共96頁2021-10-27MOM-Problems-72材料力學第5章作業(yè)/P111-5.8:SolutionThe cantilever beam shown in the figure has moments of inertia I1 and I2 in parts AC and CB, respectively. (a) Using the m

47、ethod of superposition, determine the deflection B at the free end due to the load P. (b) Determine the ratio r of the deflection B to the deflection 1 at the free end of the prismatic cantilever beam with moment of inertia I1 carrying the same load. BAL/2L/22I1ICxy(a)PB2/2LCBCxyA1B2C第72頁/共96頁2021-1

48、0-27MOM-Problems-73BAL/2L/22I1ICxyPBA2I1ICPB1IC2/PLPB2/2LCBCxyA1B2C232223248522/2/32/EIPLEILPLEILPC222222832/2/22/EIPLEILPLEILPC131312432/EIPLEILPB2113132223712424283485IIEIPLEIPLLEIPLEIPLB第73頁/共96頁2021-10-27MOM-Problems-74BAL/2L/22I1ICxyP2113132223712424283485IIEIPLEIPLLEIPLEIPLB(b)BAL1IxyP1313EIPL

49、87137124211321131IIEIPLIIEIPLrB第74頁/共96頁2021-10-27MOM-Problems-75BAL/2L/22I1ICxyPDiscussionUse energy methodLxPxM)(Lx0LLLEIdxxMEIdxxMU2/122/0222)(2)(LLLdxLxdxLxIIEIP2/22/0221122242472332112LLIIEIP211327148IIEILPWPB2121137124IIEIPLB第75頁/共96頁2021-10-27MOM-Problems-76材料力學第5章作業(yè)/P111-5.9:SolutionBAqL/2L/

50、2BAq1. 解除某些約束，施加相應約束反力，使之成為靜定基；2. 變形協(xié)調方程：RB=BAqBARBfBufBdBdBuff查表得EILRfBBu33EIqLLEILqEILqLffCCBd384726282243412873847343qLREIqLEILRBB第76頁/共96頁2021-10-27MOM-Problems-773. 求其它約束反力：1287qLRBBAqRBRAMA0200420BABAARRLqYLRLLqMML/2L/212891287128572qLMqLRqLRABA第77頁/共96頁2021-10-27MOM-Problems-78材料力學第5章作業(yè)/P112-

51、5.13*:SolutionThe bars 1 and 2 have the same tensile rigidity EA. (1) If the beam AB is considered as rigid, determine the internal forces in the two bars. (2) If the beam AB is considered as deformable with bending rigidity EI, determine the internal forces in the two bars. (1) the beam AB is consi

52、dered as rigidPLaAB21CaaAyRP1N2NAxRBACa(a) Equation of equilibrium PaaNaN221(b) Equation of compatibility 2121PNN21212第78頁/共96頁2021-10-27MOM-Problems-79(1) the beam AB is considered as rigidPLaAB21CaaAyRP1N2NAxRBACa(a) Equation of equilibrium (b) Equation of compatibility 2121PNN21212(c) Force-displ

53、acement relations EALN11EALN22得補充方程2121NN解得PN511PN522第79頁/共96頁2021-10-27MOM-Problems-8012(2) the beam AB is considered as deformablePLaAB21CaaAyRP1N2NAxRBACa(a) Equation of equilibrium (b) Equation of compatibility PNN212C2121C(c) Force-displacement relations EALN11EALN22EIaNPC48231得補充方程EALNEIaNPEAL

54、N231121482解得33121523AaILPAaILN322156AaILILPN第80頁/共96頁2021-10-27MOM-Problems-811. 危險點位于圓軸固定端橫截面的最上處和最下處（A點和B點）材料力學第6章作業(yè)/P135-6.1(d):mPLdmTMPLSolution:忽略橫力彎曲剪力的影響2. 用單元體表示危險點的應力狀態(tài)VP3. 用應力圓表示危險點的應力狀態(tài)(,)(-,)(0,-,-)B點點A點點AB332dPLWM316dmWTtA點：或B點：或第81頁/共96頁2021-10-27MOM-Problems-82材料力學第6章作業(yè)/P135-6.3(d):40

55、2040如右圖所示單元體，畫出對應的應力圓Solution maxmax(-40,-40)(-20,40)1. 主應力和主方向 3 3 1 102023.7123.1140220402204022231MPaMPa42040)40(222tan0yxxy)(02.52,04.1042)(98.37,96.75210030011.2371.232. 如圖所示在單元體上畫出主平面方向52.023. 最大剪應力23.41231max第82頁/共96頁2021-10-27MOM-Problems-83材料力學第6章作業(yè)/P135-6.4(b)*:30305020如右圖所示單元體，畫出對應的應力圓Sol

56、ution 1 1 2 2 maxmax(30,-20)(50,20)60MPa32.5260sin)20(60cos2503025030MPa66.1860cos)20(60sin25030(52.32,-18.66) =30第83頁/共96頁2021-10-27MOM-Problems-84材料力學第6章作業(yè)/P136-6.7*:Solution30如右下圖所示，在A點取單元體并進行應力狀態(tài)分析MPadP21.61002. 0052. 010203MPadTdT63.70002. 0052. 0600222/222 =120 2222222222rddddAddAdAIWAdAddAdpt

57、A test of thin walled tube is shown in the figure. If the loads are a concentrated force P=20kN and a torque T=600Nm. The diameter of the tube is d=50mm, thickness is =2mm. (1) Determine the normal and shear stresses on the inclined section at point A. (2) Determine the magnitude and orientation of

58、the principal stresses at point A. (Draw stress element.) 第84頁/共96頁2021-10-27MOM-Problems-85(2)038.4659.10763.70221.61221.6122231MPaMPa308. 221.61)63.70(222tan0yxxy)(71.56,43.1132)(29.33,57.662300100 (61.21,-70.63)(0,70.63) 3 3 1 166.5733.29(1)MPa86.451202sin)63.70(1202cos221.61221.61MPa81. 81202cos

59、)63.70(1202sin221.61第85頁/共96頁2021-10-27MOM-Problems-86材料力學第6章作業(yè)/P136-6.11:SolutionxxyyWhen a train pass through a steel bridge, the strains tested by instrument are ex=0.0004, ey=-0.00012 at point A. Find the normal stresses in directions x and y at point A. (E=200GPa, =0.3) yxxEe1xyyEe10z)(1zyxxE e

60、 e )(1xzyyE e e yxxEee21xyyEee21MPa800第86頁/共96頁2021-10-27MOM-Problems-87材料力學第6章作業(yè)/P137-6.17*:Solution薄壁上任意一點都是危險點，在薄壁上任一點A取單元體并進行應力狀態(tài)分析A MPattDPtptD6 .10015. 0185. 010200015. 04104185. 0436MPatptD7 .2403. 0104185. 026 A thin-walled tube shown in the figure is made of cast iron. Knowing its outside d