可靠性工程基础习题答案

习题一答案1.t=10536.1h；2.f(t)=00≤t＜μ)(tet≥μMTTF=1／λ+μ3.R(t)=1t＜022tet≥0λ(t)=0t＜0tt≥04.（1）由式（1-4）得：)0,0()(1)()(tetFtRmt（2）由式（1-6）得：mtetmtFtf)()()()(（3）由式（1-10）得：1)()()(mttRtft5.975.0)2000(ˆR025.0)2000(ˆF895.0)80004000(ˆR95.0)4000(ˆR05.0)4000(ˆF105.0)80004000(ˆF6.h/0033.1)20(ˆfh/002)20(ˆ7.h/109683.3ˆ5853.0)4000(ˆRt=17467.1h习题二答案1.证明：DCBACDBABACDCABBACDCABBADCADCBCBA)()()(DCBADCADCBCBA2.(1)n=2并联RS(100)=0.991；(2)n=3并联RS(100)=0.9991；(3)2/3［G］RS(100)=0.975；MTBF=833.3h3.(1)并串联：)2(1nnSRRR(2)串并联：nnSRRR)2(2(3)比较RS1和RS2大小：由于0＜R＜1n＞1的自然数当R→1时(2–Rn)→1(2–R)n→1当R→0时(2–Rn)→2(2-R)n→∝可见0＜R＜1，n为2，3···时(2–R)n＞(2–Rn)RS2=Rn(2-R)n＞RS1=Rn(2-Rn)4.。-16h1040,3,6knknttinikneeCtR)1()(15.(1)每台发动机tetR3105.0)((2)飞机为2／3［G］ttSeetR0015.0001.023)(7722.0)1()1()(3288.0288.061361eeCeeCtRiiknttiniknh23740104011)(MTBF6360inkiiidttR(3)9931.0)100(9999.0)10(SsRR6.965.0sR7.(1)选R2为Rx,当R2正常时,系统简化图(a);当R2失效时,系统简化图(b)。①图（a)为串并联系统，其可靠性为0.9312)8.01)(8.01(10.9)0.7)(1(11)1)(1(1)1)(1(1)(/3451RRRRtRSRx②图（B)为并串联系统，其可靠性为8768.0)8.09.01)(8.07.01(1)(1)1(1)(/3541RRRRtFSRx0.914880.87680.30.93120.7)(/)()(/)(tFSRtFtRSRtRRxxxxS(2)用不交MPS法：①①系统有4个MPS—411RRP532RRP5424RRRP3213RRRP②列出系统工作的MPS表达式并进行不交化4321PPPP系统工作=14351232452453112354354135114245)(133112354)35411(14245)(123)(35)(14123)53()14(35)14(148.94848.0SR9.tttSeeetR)()()(3213231)(h9.860)(MTBF0dttRS③系统可靠度)2453112354354135114()(PPRS系统工作5423132154534153141)1)(1()1)(1()1()1(RRRRRRRRRRRRRRRRRRR91488.003024.000784.01008.0216.056.010:二单元冷贮备系统有)()()1(121211ttSWteeRetR时当2199998.0)1098.11()1()(31023eetRtRtSWSh)(9950)0991(1021)1(111)2(4-21SWSWSRRm11.))(1())((43214321423433432143243142132143413221423433RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRS习题三答案1.(1)判断9.060.813求预预预预预SDCBASRRRRRR(2)分配,98.0分配AR,96.0分配BR,99.0分配CR.97.0分配DR(3)检验9.0903.0求分配SSRR2.(1)判断7.0316.0)1000(求预SSRR(2)分配,9916.0)1000(分配AR,9899.0)1000(分配BR,9923.0)1000(分配CR.9907.0)1000(分配DR(3)检验7.070005.0)1000(求分配SSRR3.-13h1068.4设备791.050）（设备SRh68.2131MTBF设备4.(1)8460.0)2()1()2(ECADBAAQQRQQRRQRR上限上限上限(2)8241.00.15950.37330.2913)2()1()1()3(下限下限下限下限ΔRΔRRR(3)0.83541)1(1)3()2(）（下限上限RRRS习题五答案1.（1）画故障树链图TACWB2B1设B1、B2组成的并联系统为W。(2)故障函数系统可靠。系统失效，设,0)(,1)(XΨXΨ最小割集(A),(B1B2),(C)。21)()(31BBcAjjxxxxXkX2121212121)()(31BBCABBCBBABBCACABBcAjjxxxxxxxxxxxxxxxxxxxxXkX2.图5-15图5-5解：（1）图5-15的故障树参照图5-5的故障树绘制图5-15的故障树（略）。(2)割集：(1,2,3,6),(1,2,4,5),(3,6,1,2),(3,6,4,5),(2,3,6),(1,3,6),(1,2,6),(1,2,3)8个。其中MCS：(1,2,3),(1,2,6),(1,3,6),(2,3,6),(1,2,4,5),(3,6,4,5)6个。(3)比较输电线的重要性(见右表)。输电浅代号在3阶MCS中出现次数在4阶MCS中出现次数重要性顺序1234563333111221111221(4)与图5–5比较可靠性大小图5-5有6个三阶MCS,而本题图5-15有4个三阶MCS和2个四阶，故本题可靠。3.解：(1)画故障树系统失效124失效1343失效12失效134失效431213失效413(2)故障函数的MCS表达式故障树的割集：(1,2,3),(4,3),(1,2,1,4),(1,2,3,4),(4,4,1),(4,4,3)MCS：(1,4),(3,4),(1,2,3)321443141111233414()(qqqqqqqqqPTP(3)求各底事的结构重要度0.3751)1(21,,324434,211st432qqqvqqqqqqqqQIii125.0)2(,212stiivqqQI375.0)3(,213stiivqqQI625.0)4(,214stiivqqQI列表求结构重要度，如对单元1，即求：?)1(stI序号134组合原割集加1变割集1000×××20014√30103401134√51002610124√711023√8111234√375.08321)1(11stnIn对单元2，即求：?)2(stI序号134组合原割集加1变割集1000×××2001430103401134√51001610114√711013√8111134√125.08121)2(21stnIn对单元3，即求：?)3(stI序号124组合原割集加1变割集1000×××20014√30102401124√51001610114√711012√8111124√375.08321)3(31stnIn对单元4，即求：?)4(stI序号123组合原割集加1变割集1000×××20013√30102401123√51001√610113√711012√8111123√625.08521)4(41stnIn习题六答案2.,)(3201,)(31206,)(3402,)(3603,)(3804,)(31005)(6.63S)(8.19012000SR用%表示：%59.1120008.1903SSR3.eRR602.0,I301.0eppIewRI2p018856.0总输出功率的容差为：%3.11113.03Ww5.(1)判断稳压器能否使用∵220v+10v＞225v＞220v-10v∴该稳压器可以使用。(2)判断稳压器能否作为合格品出厂？)V(84.216)V(16.223V)(16.3mmDA由于225V不在216.84～223.16(V)之间，∴不能作为合格品出厂。(3)不合格品出厂给社会带来的损失∵D=L(220+10)=K(220+10-220)2=100∴K=1用户损失为：L(225)=1×(225-220)2=25（元）社会损失为：25-A=25-10=15(元)习题七答案2.Z=4，R=0.94683=0.9999683，F=0.00003173.(1))N/cm(2.3,802bbsS(2)22/43.12/775.795rrLL(3))mm(10.077.6034.0,764.62017.0,382.3091.3drrdrZ直径为