机械原理作业答案 叶仲和 蓝兆辉版

REFERENCEANSWERforEXERCISEBOOKof《MechanismsandMachineTheory》TRGofMachineryTheoryandDesignCollegeofMechanicalEngineeringFuzhouUniversity2003NameClassStudentNo.Date2-1DrawthekinematicdiagramsofthemechanismsshowninFig2-1.4scale3:11BC23A42A3C1BFig2-1(a)11223344ABCABCFig2-1(b)NameClassStudentNo.Date2-2DrawthekinematicdiagramsofthemechanismsshowninFig2-2.1233445566ABCDEABCDEFF21Fig2-2(a)NameClassStudentNo.Date2-3DrawthekinematicdiagramofthemechanismshowninFig2-3.8I7B65HA1G2C3E4DF278I156HAGFCB3ED4Fig2-3NameClassStudentNo.Date2-4CalculatethedegreeoffreedomofthemechanismsshowninFig2-4,andpointoutwhatshouldbepaidattentiontoduringthecalculation.解：(a)左右为对称结构，设左侧为虚约束。(b)E为杆4、5、6的复合铰链。(c)滑块7与机架8间为移动副。F=3n-2PL-Ph=3×7-2×10=1解：(1)红线内的构件为重复结构，构成虚约束。(2)去掉以上构件后，C仍为构件2、3、4的复合铰链。(3)滑块5与机架6之间为移动副。F=3n-2PL-Ph=3×5–2×7=1ABCDEFGHIJRedundntconstraint2345678Fig2-4(a)1BCDEFGHIJKRedundantconstraintA23456Fig2-4(b)NameClassStudentNo.Date2-5CalculatethedegreeoffreedomofthemechanismsshowninFig2-5,andpointoutwhatshouldbepaidattentiontoduringthecalculation.解：(a)两个滚子有局部自由度。(b)滚子D与凸轮1之间只能算一个高副。F=3n-2PL-Ph=3×7–2×9-2=1解：(1)杆件BC与齿轮2焊接在一起。(2)A为齿轮4、杆件1和机架5的复合铰链。F=3n-2PL-Ph=3×4–2×5-1=1常见错误：认为B是复合铰链，而不认为A是复合铰链。ABCDEFGLMNO12345678Fig2-5(a)BDA51234Fig2-5(b)NameClassStudentNo.Date2-6CalculatethedegreeoffreedomofthemechanismsshowninFig2-6,andpointoutwhatshouldbepaidattentiontoduringthecalculation.解：(a)C为构件2、3、4的复合铰链。(b)C处有两个转动副和两个移动副。E处有一个转动副和两个移动副。F=3n-2PL-Ph=3×7–2×10=1注意：E不是复合铰链！解：当构件尺寸任意时，构件2作平面复杂运动，而杆4与机架间组成移动副，所以杆4仅作平动。因此，构件2和构件4之间有相对转动。因此，应该有构件6，并且构件4和6之间有转动副，如右图所示。当AB＝CD且BC＝AD时，杆2仅作平动。杆4与机架间组成移动副，所以杆4也仅作平动。这样，构件2和构件4之间就没有相对转动，只有相对移动。即：构件4和构件6之间就没有相对转动了，因此，可将构件6与构件4焊接起来（去掉构件6），如左图所示。然而，在计算机构自由度时，应该按一般尺寸情况下进行分析，即：应该按照右图情况来分析机构的自由度。F=3n-2PL-Ph=3×5–2×7=1ABCDE1236784Fig2-6(a)ABCDAB=CDBC=ADABCD12345123456Fig2-6(b)NameClassStudentNo.Date2-7ThekinematicdiagramofanenginemechanismisgiveninFig2-7.(1)Calculatethedegreeoffreedomofthemechanism,andpointoutwhatshouldbepaidattentiontoduringthecalculation.(2)Makethestructuralanalysisforthemechanism.(3)MakethestructuralanalysisforthemechanismwhenlinkEFGisregardedasthedriver.Note:Duringstructuralanalysis,listtheassemblyorderofAssurgroups,thetypeofgroup,thegradeofgroup,thegradeofthemechanism,thelinkserialnumbers,theinnerpairandtheouterpairsofeachgroupineachmechanism.解：（1）F=3n-2PL-Ph=3×7–2×10=1（2）当AB为原动件时，（3）当EFG为原动件时，ABCDEFGH12346578第一杆组RRPRRR2,34,5内副E外副B,移CF,D第二杆组类型杆号第三杆组RRP6,7转H6-7G,移H转C2-33-87-8ABCDEFGH12346578III级杆组RRP1,2,3,46,7内副外副A,E,移C类型杆号B,D,转C3-8转HG,移H7-8NameClassStudentNo.Date2-8MakethestructuralanalysisforthemechanismshowninFig2-8.(a)Whenlink1isregardedasthedriver.(b)Whenlink5isregardedasthedriver.Note:Duringstructuralanalysis,listtheassemblyorderofAssurgroups,thetypeofgroup,thegradeofgroup,thegradeofthemechanism,thelinkserialnumbers,theinnerpairandtheouterpairsofeachgroupineachmechanism.解：(a)当1为原动件时杆件2,3,4和5组成一个三级Assurgroup.(b)当5为原动件时杆件3和4组成第一个RPRAssurgroup.杆件1和2组成第二个RPRAssurgroupBC3621AD45EFig2-8236A1CDB54E2361ACBD5E4ABCDEFABCDEFABCgradeIIgradeIIIgradeIV236A1CDB54EBD36A1524CECADBD53B4A6CENameClassStudentNo.Date2-9TheschematicdiagramofapunchmachinedesignedbysomeoneisshowninFig2-9.Thismachineshouldbeabletotransformacontinuousrotationofgear1intoatranslationofthepunch4.Canthemachineworkproperly?Ifitcan’t,pleaserectifyit.解：不能正常工作。改正如图(或者改成题目2-3构件5、6、7的连接)12345Fig2-9NameClassStudentNo.Date2-10TheschematicdiagramofamechanismdesignedbysomeoneisshowninFig2-10.Thismechanismshouldbeabletotransformacontinuousrotationoflink1intoanoscillationoflink4.Canthemechanismworkproperly?Ifitcan’t,pleaserectifyit.解：不能正常工作。改正后ADBCE12345Fig2-10ADBCE124ABCEDABCNameClassStudentNo.Date3-1LocateallinstantcentresofmechanismsinthepositionshowninFig3-1.2134P14P12P23P34P13P24Fig3-1(a)1234P12P24∞)P14P23P34P13∞Fig3-1(b)12123P23∞PP13OnnFig3-1(c)2414P(2A115(P)GB25(P)nP()12E(P45)PJHFn423CP(3D))P34()122441214PPP14PP15P451545Fig3-1(d)NameClassStudentNo.Date123ABCEDGF12P()13P()P()P23P14∞V1()Fig3-1(e)1233411413P12PP23PP43P132324P12P21P14P42PP243434AP14)(BP12CP()2334(PD)1234∞P1324PFig3-1(f)NameClassStudentNo.Date3-2InthepositionshowninFig3-2,determinetheratio3/1oftheangularvelocityofgear3tothatofgear1,usingthemethodofinstantcentres.解：P13是构件1和3的瞬心，等速重合点，所以1LAE=3LDE3/1=LAE/LDE3-3InthepositionshowninFig3-3,determinetheratio2/1oftheangularvelocityoffollower2tothatofcam1,usingthemethodofinstantcentres.解：E（P12）是构件1和2的瞬心，等速重合点，所以1LOE=2LAE2/1=LOE/LAE612345ABCDEF123361PPPPPP161312236313()P36()P23()P12()P16E(13P)Fig3-2OA()()T123121323PPPE()Fig3-3NameClassStudentNo.Date3-4Inthepivotfour-barlinkageshowninFig3-4,1=10rad/sec.Usingthemethodofinstantcentres,(a)FindthevelocityofpointCinthepositionshowninthefigure.(b)Inthepositionshowninthefigure,locatethepointEonthelineBC(oritsextension)whichhastheminimumvelocityamongallpointsoflineBCanditsextension,andthencalculateitsvelocity.(c)drawtwopositionsofthecrankABwhenVC=0.解：（a）VB1=1LAB=VB2=2LFB，所以VC=VC2=2LFC=1LABLFC/LFB(b)VE=2LFE。(c)VC=0所对应的曲柄AB的两个位置:NameClassStudentNo.Date14A(P)12PB()23C(P))DP(341234P24Eω