0.+ 所以 04s b=-45 214.22 74.22 G1=21422MPa,a,=0,σ3=-7422MPa 4.图示封闭薄壁圆筒,内径d=100mm,壁厚t=2mm,承受内压p=4MPa, 外力偶矩M=0.192kN·m。求靠圆筒内壁任一点处的主应力 0.192×103 解: (0.1042-01)x005=575MPa a.=P=100 MPa Omax=ox+ox+.ox-o 100.7 MPa 2 2 49.35 100.7MPa,a2=49.35MPa MPa 5.受力体某点平面上的应力如图示,求其主应力大小。 解:取坐标轴使a4=100MPa,r1=20MPa 40 MPa 100 MPa ,+0,O、-0 20 MPa 2cos 2a-T sin 2a 100+.100-0 cos120°-20sin120°=40 得σn=43.MPa 106.33 )2 MPa 1=10633MPa,a2=36.77MPa,a3=091 所以 2 2 min max ) 2 ( 2 xy x y x y        + −  + = 74.22 214.22 − = MPa  1 = 214.22 MPa， 2 = 0， 3 = −74.22 MPa 4. 图示封闭薄壁圆筒，内径 d =100 mm，壁厚 t = 2 mm，承受内压 p = 4 MPa， 外力偶矩 Me = 0.192 kN·m。求靠圆筒内壁任一点处的主应力。 解： 0.05 5.75 32 π(0.104 0.1 ) 0.192 10 4 4 3  = −   x = MPa 50 4 = = t pd  x MPa 100 2 = = t pd  y MPa 49.35 100.7 ) 2 ( 2 2 2 min max + = −  + = xy x y x y        MPa 1 =100.7 MPa， 2 = 49.35 MPa， 3 = −4 MPa 5. 受力体某点平面上的应力如图示，求其主应力大小。 解：取坐标轴使  x = 100 MPa， x = 20 MPa          cos 2 sin 2 2 2 x x y x y − − + + = cos120 20sin 120 40 2 100 2 100 − = − + + = y y     得  y = 43.1 MPa 2 2 min max ) 2 ( 2 xy x y x y        + −  + = 36.77 106.33 = MPa 1 =106.33 MPa， 2 = 36.77 MPa， 3 = 0  x y      = −45  45  45  45 Me Me t p 40 MPa  120 20 MPa 100 MPa