解:E=E,=0 (a,+E +-(8r-E)cos 2a+-yu sin 2a=-yu sir 2 sin(90°+2a)E g′5.00×10-4 tan 2a= 3.75×10-43 sin 2a=0.8 e 28 M=n pT WG sin 2a 16 2(1+v) sin 2a 兀×0.13×200×10°×2×500×10- 一=1.96×10N.m=196kN.m 16×2×1.25×0.8 54由电阻应变计法测得钢梁表面上某点处c=500×10,E,=-465×10,已知 E=210GPa,v=0.33。试求σ,及σ,值 E 解: e va 210×10 1-0.33|500×10-6+0.33×(-465×10)=81.7MPa 210×10 465×10+0.33×(500×10-°)=-70.7MPa 1-0.33 55有一处于平面应力状态下的单元体,其上的两个主应力如图所示。设 E=70GPa,v=0.25。试求单元体的三个主应变,并用应变圆求出其最大切应变ymxo R, vo 80 80 MP: E-E=70×10(-023)=8.37×10 E2=8.57×10 E2x(0.25×80×1 Vo VO2 )=-5.72×10 70×10 由应变圆可知: y 2 =8.57×10-+5.72×10+=14.29×10 H [ H \  H HD H H H H D J D J VLQ D   VLQ      FRV        [  \  [  \  [\ [\ c VLQ      H H J D D cc   $ $ [\ H H D D cc c VLQ   VLQ  $       WDQ    u u cc c   H H D VLQ D  S H S : 0 : 7 W D H W J VLQ   c ** [\ D H D Q H W VLQ   VLQ       H S S c  u c G ( 0 : : *   1 P               u u u u u u u u u u   N1P           H [ u H \  u ( *3DQ  V [ V \ ° ¯ ° ® ­             \ \ [ [ [ \ ( ( H QH Q V H QH Q V >      @ 03D         u  u  u  u   V [ >      @ 03D          u  u u   u   V \  (  *3DQ  PD[ J ( (    V QV H           u u    u     H u ( (    QV QV H            u u u u      u   PD[   J H  H    PD[            J  HH u  u u 03D 03D