工程力学2习题课

1、CABD600300500400E40kN55kN 25kN20kNCABD600300500400E40kN55kN 25kN20kNCABDE40kN55kN 25kN20kNRkN10100 020202525555540400 0 RRFxCABDE40kN55kN 25kN20kN10 01 1 RFN)()(RF kNN10101 1 R40kNFN220kNCABDE40kN55kN 25kNR20 040402 2 RFN)()(RF kNN505040402 2FN320kN25kNCABDE40kN55kN 25kN20kNR30 0202025253 3 NF)()kN(

2、N 5 53 3F20kNFN440kN55kN 25kN20kNR4)()(F kNN20204 45010520+CABD600300500400E40kN55kN 25kN20kN)(FkNNmax5050 F1F2F3l1l2l3ABCDF1F2F3l1l2l3ABCDF1FN1)(kNNN 20200 01 11 11 1FFFF2F1FN2F1F2F3l1l2l3ABCD)(kNNN 15150 02 22 22 21 1FFFFRFN3)(kNNN 50500 03 33 3FRFFN2 =-15kN （-）FN1 =20kN （+）FN3 =- 50kN （-）15+-2050

3、F1F2F3l1l2l3ABCDFN2 =-15kN ( - )FN1 =20kN （+）FN3 =- 50kN ( - )F1F2F3l1l2l3ABCD)(MPa.N 8 81761761 11 1AFAB )(MPa.N 6 674742 22 2AFBC )(MPa.N 5 51101103 33 3AFDC (F1F2F3l1l2l3ABCDm102.534-N 1 11 11 1EAlFlABm101.424-N 2 22 22 2EAlFlBCm101.584-N 3 33 33 3EAlFlCD-0.3mm BCCDBllumm10-0.474- CDBCABADllllMex

4、nnMeMexMexnPM9549e mN6379mN4780mN159004e3e2e1e MMMM 0023e2e TMMMx mN95603e2e2 MMTmN47802e1 MTxMe2Me3T2Me21T133Me4Me1Me3Me2Me2mN47802e1 MTmN63704e3 MTmN9560max TD+M2M3M1180)2(pepe GIaMGIaM3m4mABCDE4m4mRARBF1=2kNm=10kN.mF2=2kNF1=9kNF2=4kNACBD1m1m1mbhzy四、计算 309090tc1. 构件内危险点应力状态如图所示，试作强度校核：1）材料为铸铁，已知许用

5、拉应力MPa,压应力MPa，泊松比 =0.25；2）材料为铝合金，MPa；3）材料仍为铸铁，应力分量中为压应力。15MPa15MPa 2213131151154 1522024.330t 1，r1）求主应力 （-9.27）MPa铸铁为脆性材料，且，主应力中拉应力占优，主要是最大拉应力引起脆性断裂，选第一强度理论 MPa解：MPa 13222222212233122224.279.2733.54904154 1533.5412124.279.2724.279.27230.090 334rrr2)铝合金为塑性材料，选第三强度理论 MPaMPa或 MPa选第四强度理论 MPaMPa 22223153 1530.090 4r 或 MPaMPa 221 313123151154 152209.270