(7-0.28×77)=-0.07068×10 206×10 所画的圆变成椭圆,其中 d1=d+d1=300×(1+0.364×10-)=30109mm(长轴) d2=d+d2=300×(1-0.07068×10-)=299979mm(短轴) 7-14已知一受力构件表面上某点处的ax=80MPa =-160MPa,a=0,单元体三个面上都没有切应力。试求该点 160 MPa 处的最大正应力和最大切应力。 =80 MPa 80 MPa 1=80MPa,a3=-160MPa 120 MPa 2 15单元体各面上的应力如图所示。试用应力园的儿何关系求主应力及最大切应力 解:(a)由xy平面内应力值作a,b点,连接φb交σ轴得圆心C(50,0) 应力圆半径 )2+402=447 T/MPa 30MPa 故a1=50+447=947MP G3=50-44.7=53MPa 40MP g/ MPa 3=44.7MPa 2 (b)由xz平面内应力作a,b点,连接 mb交σ轴于C点,OC=30,故应力圆半 T/MPa 50 G1=30+50=80MPa 40MPa o,=50MPa t, 50MPa 2 (c)由图7-15(c)yz平面内应力值作a T/MPa b点,圆心为O,半径为50,作应 力园得 80MP 2=-50MPa /MPa l09                   u  u u H QVV (      PP     u  u  G G GH      PP     u  u  G G GH  03D V [ 03D V \ V ]  03D V PD[ V [ 03D V 03D V    03D       PD[   V V V  D [\ D E DE V &              U   03D V    03D V   03D V  03D    PD[  V V W E [] D E DE V & 2&       U    03D V  03D V  03D V   03D     PD[   W F  F \] D E 2  03D V 03D V   03D V    03D V [  03D V \ 03D 03D 03D 03D [ \ ] V V R      E D W PD[  W 03D V 03D 03D 03D [ \ ] 03D V V  R  PD[ W    E D  V  W  03D V 03D 03D 03D [ \ ] V V  R  E  V  PD[ W D  W 03D V 03D