西安建筑科技大学：《材料力学》课程习题与答案_第12章 动载荷与交变

第26次作业: 6-2 2×1020 d=1+-=1.23 ,4“个P=665KNm M 665×10 W2×237×103 140.3MPa<[a] 6-6 =314 3142×0.2 =26.15 98 q4=Kq=26.15×7.75×98×0.056×0.028=3l1kN/m 9=156kNm 6×1.56×10° =107MPa 28×56 6-9 PP5000×16003 b=3E/3×210×103×2500×10+=1.3mm 2×10 K,=1+|1+ 1+,1+ =5.05 Odm=kO,=505×500069=1616MPa 250×103 6-14 P64×2000×30003 21.11mm 3E3×11×103×3.14×200 2.2 Vg△nV9.8×0211 32×2000×3000 Dmx =Ko,=2.2 =16.8MPa 3.14×200

第 26 次作业： 6-2 140.3 [ ] 2 237 10 66.5 10 66.5 . 4 4 1 1.23 9 20 3 2 10 3 6 max max 1 2 max 3  =      = = = + = = + = =  = MPa W M k N m P l K Pl M g a K a d d 6-6 MPa W M k N m q l M q K q k N m K n d d d d 107 28 56 6 1.56 10 1.56 . 8 26.15 7.75 9.8 0.056 0.028 3.11 / 26.15 9.8 31.4 0.25 1 31.4 60 2 2 6 max max 2 max 2 =    = = = = = =     = =  = + = =    6-9 K MPa h K mm EI Pl d d st st d st 161.6 250 10 5000 1600 5.05 5.05 1.3 2 10 1 1 2 1 1 1.3 3 210 10 2500 10 5000 1600 3 ,max 3 3 4 3 3 =   = =  =  = + +  = + + =       = =   6-14 K MPa g v K mm EI Pl d d st st d st 16.8 3.14 200 32 2000 3000 2.2 2.2 9.8 0.02111 1 21.11 3 11 10 3.14 200 64 2000 3000 3 ,max 3 2 3 4 3 3 =    = =  =  =  = =        = =  

6-16 -20 △a=60-(-20)=80MPa (b)r =-3△a=20-(-60)=80MPa 20 =2△a=-30-(-60)=30MPa 0-∞△G=0-(-60)=60MPa

6-16 r MPa r MPa b r MPa a r MPa 0 ( 60) 60 0 60 2 30 ( 60) 30 30 60 3 20 ( 60) 80 20 60 ( ) 60 ( 20) 80 3 1 60 20 ( ) = −  = − − = − = =  = − − − = − − = = −  = − − = − = = −  = − − = − =    