解析法 46. IMPa maX =(a+a)±VGx-a,) 2 2 1)+4 26.IMPa min 2 Mo==tg x=1685° 2 o46.1MPa,2=29MPa,3=-26.1MPa 01-O maX 236.1MPa解析法： o x y x x y x y x t g MPa MPa 16.85 2 2 1 26.1 46.1 ( ) 4 2 1 ( ) 2 1 1 0 2 2 min max = − − = − = +  − + = −             1 = 46.1MPa,  2 = 29MPa,  3 = −26.1MPa 36.1MPa 2 1 3 max = − =   