《材料力学》课程教学课件（PPT讲稿）第七章 应力应变分析及强度理论 Analysis of Stress and Strain Failure Criteria §7-2 平面应力状态分析-解析法（Analysis of plane stress-state）

应力应交状态分析(Analysis of stress-state and strain-state)R将α代入正应力公式arctanO+0O117cos2αo-Tx, sin2αoa22十ODd+ cos2αotan 2αoTH22O福十X...tan2αo果x122/tan*2α+2T十OOxyXX222Txy+O

(Analysis of stress-state and strain-state) 将 0 代入正应力公式 0 0 0 cos 2 sin 2 2 2 x y x y  xy         + − = + − 0 arc   = −    −   xy x y 1 2 tan 2     0 0 cos 2 tan 2 2 2 x y x y xy        + −   = + −     0 0 1 2 2 2 2 1 x y x y xy        + −   = + −   +   2 tan tan 1 2 2 1 x y x y xy      + −   = + +   −       + −   xy x y xy x y 2 2 2      

应力应交状态分析(Analysis of stress-state and strain-state2T+OOxyX2221C+122txy2T0十X1+222+12txy+12

(Analysis of stress-state and strain-state) 1 2 2 1 x y x y xy      + −   = + +   −       + −   xy x y xy x y 2 2 2       ( ) 1 1 2 2 1   x y   x y     + − −   +     = + +       −       + −   xy x y xy x y xy x y 2 2 2 2 2 2          ( ) 1 2 2   x y +   x y −   = + +   −   xy x y 2 2  