材料力学2013典型例题

1、F1F2F3l1l2l3ABCDF1F2F3l1l2l3ABCDF1FN1)(kN2001N1N1 FFFF2F1FN2F1F2F3l1l2l3ABCD)(kN1502N2N21 FFFFFFN3)(kN5003NR3N FFFDFN2 =-15kN （-）FN1 =20kN （+）FN3 =- 50kN （-）15+-2050F1F2F3l1l2l3ABCDFN2 =-15kN ( - )FN1 =20kN （+）FN3 =- 50kN ( - )F1F2F3l1l2l3ABCD)(MPa.N 8 81761761 11 1AFAB )(MPa.N 6 674742 22 2AFBC )(M

2、Pa.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- CDBCABADllllD+Me2Me3Me1180)2(pepe GIaMGIaMMPa7 .69tmaxmax WT mkN292e M33. 2180)23(pepe GIaMGIaMCBBACA D+Me2Me3Me2RRqlFFBA lqFFABx)0(222)()0(2)(2RRSlxqx

3、qlxxqxxFxMlxqxqlqxFxFAA )0(2)(SlxqxqlxF2SqlF 2SqlF +ql/2ql/2BlqFAxF)0(222)(2RlxqxqlxxqxxFxMA lqFABxF00 Mx，0, Mlx( )d0d2M xqlqxx2lx 822maxqlMMlx+82qll/2 lqFABxF+ql/2ql/2+82qll/282maxqlM2maxSqlFACBFa2a203014FRAFRBFa433cm07. 112)cm2)(cm4 . 1(12)cm2)(cm3( zI34maxcm07. 11cmcm07. 1 yIWzzmaxWMz kN3 aWFWFaz

4、zmax maxwABqRR2ABqlFFABqFRAFRBx2( )22qlqM xxx2346qlqEIwxxC 222qlqEIwxx 341224qlqEIwxxCxD233(64)24qlxxlEI 233(2)24qxwlxxlEI 0w xABqFRAFRB A B3max24ABqlEI wmax4max25384lxqlwwEI FFMPa7 .351 . 0700016MPa37. 6101 . 05043t32N WTAF32397 .35)237. 6(237. 6)2(22222minmax MPa32, 0,MPa39321 1 ABCD1.4m0.6m15kN10kN0.8mABFemkN15 6 . 0104 . 115e MkN25 FABFMe+15kNm-20kNmmkN20max MmkN15 T)1(3243 DWMPa26.157223r WTMx8801000yzyxz880FFlxz880 x8801000yzykN6 .134)11(105 . 6101 . 214. 3)(2811222cr lEIF x88