第七章应力状态和强度理论 7-1试从图示各构件中A点和B点处取出单元体并表明单元体各面上的应力。 -4F 解:(a)o=d = 4 πd 33 (b)t=16M3 w。πd3 = 16×8×103 M. =12 KN-m =8kN.m M, 4KN.m π×803×10-9 T=79.6MPa =79.6×10°Pa =79.6MPa (c)∑MA=0 12KN 0.8kN-m FB×1.20-0.8-2×0.4=0 FB=1.333kN M=F0.3=0.4knm 30 Fs=-1.3333kN 400400400 = 1.333×103 2KN TA=max2bh240×120×10 .8kN-m 6 B =0.417×10a=0.417Ma M 0.4×103 400400400 A B=140×1203×102×30×10 12 =41.7MPa =2.08×106Pa=2.08a S2=40×30×45×10-9=54×10-6m3 -TB=0.312MPa 40×120 2= 12×10-12=5.76×10-6m4 =2.08MPa FS1.333×103×54×10- TBb240×10-3×5.76×10 =0.312×10°Pa=.312MPa 6 M39.3 (d)== Wπ×20 32×10-9 =50.0×10°Pa=50.0MPa Am-8-393Nm N.m =78.6 Wπ×20×10-9 0.5m Im 16 =50.0×10Pa=50.0MPa 102  $ %  D   G  ) V E   S   G 0 : 0 W          u u u u   3D  u 03D F ¦ 0 $  )% u      u  )% N1 0 )% u  N1 P )6 N1 V $    6 PD[           u u u u EK ) $ W W   3D 03D  u               u u u u u \ , 0 ] V %   3D 03D  u          P   6] u u u u        P      u u u , ]   3D 03D              u u u u u u u    ] % ] % E, ) 6 W G        u u :] 0 V   3D 03D  u   S      u u : 7 W   3D 03D  u O   $ ) ) ) G O   O    G  ) V $ D D D   0  N1 P 0  N1 P 0   N1 P $ $ W 03D     N1P N1 $ % R ] \  N1P N1 $ )$ % )%    $  03D $ W % 03D % W 03D V % $ 0 1P P P 0 1 P