1、xylqM xqlxqx( ) 222222qlqEIyxx 2346qlqEIyxxC 341224qlqEIyxxCxD000 xyxly时，时，CqlD 3240,xqlxyABqEIlxxl2464233()233(2)24qxylxxlEImax ABqlEI3244max25384lxqlyyEI xqlxyABxylPABM xP lx( )() EIyPxPl 22PEIyxPlxC 3262PPlEIyxxCxD00,0 xyy时，CD 0 xylPABxPxEIxl22()2(3 )6PxyxlEImax BPlEI223max3BPlyyEI xylPABxxyl2PAB
2、Cl2ACM xPx段：( ) 22PEIyx 24PEIyxC 312PEIyxCxD00 xy时，D 002lxy时，CPl 216xyl2PABCl2xPEIxl16422()22(43)48PxyxlEImax ABPlEI2163max248lxPlyyEI xyl2PABCl2xyaqABCxaaaDEMxqaxxa11110()()112222()2EIyqaxqEIyqaxxayaqABCxaaaDEqaqax1x2Mxqaxqxaaxa22222222()()()121212,xxayyyy时21112qaEIyxC得CCDD1212110,0 xy时得 D10222,0 x
3、ay时得 Cqa23116 11EIyqax2222()2qEIyqaxxa3111 116qaEIyxC xD232222()26qaqEIyxxaC34222222()624qaqEIyxxaC xD22111233222223111134322222(113)06 3()1126(11)06 4()44224qaaxxaEIqaxxaaaxaEIqaya xxxaEIqyaxxaa xaxaEI 13max10116AxqaEI 24max22198xaqayyEI Af2l2l02lx ( )M xP x EIyP x 212PEIyxC 3116PEIyxCx D2lxl( )M x
4、P x 2EIyP x 2222PEIyxC 32226PEIyxC xD xl 时,y=0, =02ClxC左C右左C右时,y=y,=232211,23CP lDP l 231153,1616CP lDP l 3231153()61616yPxPl xPlEI0 x 3316APlfEI CABf、Cf53844qlEIPlEI348mlEI216AqlEI324PlEI216mlEI3BqlEI324PlEI2166mlEIEIpafc331EIpac221223cBfaPalaEI23BPalEI 21cccfff21ccclm2xm1xl/ 3EImm12/( )EIyM x 2121
5、mmlm2xm1M 0Mm2m10y 45 (2 )384CqafEI PaaEI()2162 0Pqa56BPaEI 22maEI20mPa4445 (2 )50,38424CDqaqaffEIEI 434(2 )(2 )14833BqaqaaqafEIEIEI 342(2 )82483BDfqaaqafEIEI323(2)2(2)BPaPa afEIEI33BBCfaPafEI 3512PaEIBPaEIPa aEI22 22()234PaEI顺时针 332PaEI解：解：3364BCqaqaEIEI顺时针BqaaEIqaaEI22223216()312qaEI顺时针 445824BCqa
6、qafEIEIaP=qaP=qam=q /2ABqCBP=qaqABCaaaBqkqlEIqlEI8224222433378384qqakEI顺时针 4516768CqlqlfkEI qlk88qlB处反力=kEIa23 41332CqaqafkEI 342(2 )243CqaqafaEIEI 41283CCCqafffEIBqa处 反 力 =2()BPaaPaEIEI23()22BPaaPafEIEIP alE A33CPafEI32xBPaCfEIyCBBCCfll343PaPaEIEA0Bf 即R lEIqlEIB34380RqlB380A即M lEIqlEIA32403MqlA182解
7、：解：(1)变形协调条件为：变形协调条件为：DDABCDff即5633333PaEIR aEIR aEIDD54DRP(2)(3)自行完成！自行完成！PBACDRDD如何得到？如何得到？BBABBCff334833BBR aR aqaEIEIEI即316BRqa2L2L2L2LCRCDClf3()48CCRaPRLEAEI3348CA LRPA LI a140MPa15 0 0fL2L2L31430ZWcm432240ZIcm80.4qkg m2max48P Lq LM32(50 10 ) 10(80.4 9.8) 10483134.8 10 N m2L2LmaxmaxZMW36134.8 1
8、094.31430 10PaMPa140MPaPqCCCfff3454 83 8 4ZZP Lq LE IE I3349898(50 10 ) 105 (80.4 9.8) 1048 (200 10 ) (32240 10 )384 (200 10 ) (32240 10 )317.76 1017.76mmm1500fL1020500mmmCf f设：开关闭合其状态为设：开关闭合其状态为“1”，断开为断开为“0”灯亮状态为灯亮状态为“1”，灯灭为，灯灭为“0” 0 0 0 0 C 0 0 1 10 1 0 10 1 1 01 0 0 11 0 1 01 1 0 01 1 1 156例例1：化简
9、化简CABCBACBAABCY)()(BBCABBACCAAC 例例2： 化化简简CBCAABY)(AACBCAABCBACACABABCAAB例例3： 化简化简CBACBAABCYABCCBACBAABCACBC CBCBA)(CBACBAY例例4： 化简化简BABAACBCBACBABAAB例例5：化简化简DBCDCBADABABCYDBABCDCBAABCDBCDCBAABDBCDCBAB)(DCBCDABCDBCDAB)(DADBCDCBAABCBCDABCDB58ABC001001 11 101111ABCCABCBABCAY用卡诺图表示并化简。用卡诺图表示并化简。解：解：1.卡诺图
10、卡诺图2.合并最小项合并最小项3.写出最简写出最简“与或与或”逻辑式逻辑式59ABC001001 11 101111解：解：三个圈最小项分别为：三个圈最小项分别为：合并最小项合并最小项ABCCBAABCBCACABABC BCACABABACBCY6000ABC1001 11 101111解：解：CACBYAB0001 11 10CD000111101111DBY CBABCACBACBAY(1)(2)DCBADCBADCBADCBAY61解：解：DBAYAB0001 11 10CD000111101DBDBCBAAY11111111162Y = Y2 Y3= A AB B AB.A B.A
11、B.A. .A BBY1.AB&YY3Y2.63反演律反演律反演律反演律64ABY001 100111001=A B65.A B.Y = AB AB .AB.BAYA B = AB +AB66=A B =1ABY逻辑符号逻辑符号=A BABY001 10010011167Y&1.BA&C101AA=AC +BCY=AC BC 设：设：C=1封锁封锁打开打开选通选通A信号信号68Y&1.BA&C011设：设：C=0选通选通B信号信号B=AC +BCY=AC BC69 开工为开工为“1”，不开工为，不开工为“0”； G1和和 G2运行为运行为“1”，不运行为，不运行为“0”。700111 0 0
12、1 0 100011 0 11 0 10 0 1 0 1 0 0 1 1 1 0 0 1 1 01 1 10 0 0A B C G1 G271ABCCABCBABCA1 GABCCBACBACBA2 GABC001001 11 101111ACBCAB1 G1 0 10 0 1 0 1 0 0 1 1 1 0 0 1 1 01 1 10 0 00111 0 0 1 0A B C G1 G2 100011 0 172ACBCAB1 GACBCAB ABCCBACBACBA2 GABCCBACBACBA2 G ABC0010011110111173A BCA BC&G1G274 750 0 01 0 0I0I1I2I3I5I6I输入输入输输 出出Y2 Y1 Y076Y2 = I4 + I5 + I6 +I7 = I4 I5 I6 I7.= I4+ I5+ I6+ I7Y1 = I2+I3+I6+I7 = I2 I3 I6 I7. . .= I2 + I3 + I6+ I7Y0 = I1+ I3+ I5+ I7 = I1 I3 I5 I7.= I1 + I3+ I5 + I77710000000111I7I6I5I4I3I1I2Y2Y1Y078
