解:1)计算惯性半径 41×10 二 A 5225.05mm 7.40×104 =11.58mm 522 2)计算柔度 A.= py2_0.6×580 e (O 哲 5.05 1=75 =68.9 A=522mm2 .=241 l11.0×750 l,=141×104mm 委出 =740×104n mm Zz11.58 64.8解： O l =580 l =750 12 22 6 6 24 x x y y z 1 2 O Iz=7.40×104mm4 Iy=1.41×104mm4 A=522mm2 A I i y y = 4 1.4110 = 522 = 5.05mm A I i z z = 4 7.4010 = 522 =11.58mm 1)计算惯性半径 2)计算柔度 y y   = l2 iy 0.6 = 580 5.05 = 68.9 z z   = l1 i z 1.0 = 750 11.58 = 64.8