辽宁工业大学：《材料力学》课程教学资源（作业试题）第20次作业

第20次作业: 9-2 (a)ul=5m,(b)u=0.7×7=4.9m,(c)ul=0.5×9=4.5m (d)ul=4×2=4m,(e)u=8m,(a)U=0.5×5=2.5m, (f)最大4e)最小 F=2Fn(0s30°+Fa2(0)cos30°+x2E_,E 9-10 2E|x2×190×103 (1)p= 220 (2)2-TE 215×10=65 5.8 490 (3)=、x 11×103 =73.6 第21次作业: 1=21.m=,3500 =2166 1155 2800 (2166) 2.5 F F=0928kN 3.5 σ==0.58MPa<a]=0.6MPa A 9-17 弯曲强度: M=2IFIKN m=Iow. 2351 -×100×1802=6345kN [F]=952kN 稳定性条件

第 20 次作业： 9-2 ( )最大，( )最小 ( ) 4 2 4 ,( ) 8 ,( ) 0.5 5 2.5 , ( ) 5 ,( ) 0.7 7 4.9 ,( ) 0.5 9 4.5 f e d l m e l m a l m a l m b l m c l m =  = = =  = = =  = =  =       9-7 2 2 2 3 2 2 1 2 cos 30 36 (0.7 ) 2 cos30 2 l EI l EI l EI F F F o cr o cr = cr + = + =   9-10 73.6 20 11 10 (3) 65.8 490 215 10 (2) 92.3 220 190 10 (1) 2 2 3 2 2 3 2 2 3 =   = = =   = = =   = =             P P P P P P E E E 第 21 次作业： 9-12 ( ) MPa MPa A F F F k N i mm N N 0.58 [ ] 0.6 0.928 3.5 2.5 0.06 216.6 2800 ( ) 216.6 11.55 2500 11.55 , 12 40 2 = =  = = = = = = = = =       9-17 弯曲强度： F k N k N m W k N m F M z [ ]' 95.2 100 180 63.45 . 6 1 2 235 . [ ] 3 2[ ]' 2 = = =  =    = 稳定性条件

1000 200F-T EI 15.5kN [F]"= 3 =155kN [F]=15.5kN

F k N k N n F F k N l EI F st cr cr [ ] 15.5 15.5 3 [ ]" 200, 15.5 5 1000 2 2 = = = = = = =  