g-2150e2w=li0w0x10 N 60x10 3×10 +m7.3000x10+100x10x=0.2+0273=0473≤1 所以强度满足要求 (2)弯矩作用平面内稳定性验算 4=An=150×200=30000mm W=形.=名x150x202=1000x10m 4-5×200=5774mm 1=4=23500 43.30<91 1,57.74 0=1+a1651+433016T=0693 由构件初始偏心距e。=0,得k=0 K=Neo+M 3x105 =0.188 防.+1wx1ox+\s00 60×103 pm=(1-K)(1-kK)=(1-0.188)2=0.659 60×103 m,40693x0.659x30000=438N/mm<10N/mm 弯矩作用平面内稳定性满足要求 (3)弯矩作用平面外稳定性验算 -后150=83 ，--250-57.74<91 1,43.3 1 +/651+5774165=050 4×2500x095x20-0.350 ΓVπ×1502×220 g=+1/ 2c. 2c C221 2 3 3 150 200 1000 10 6 W mm n =   =  3 6 3 60 10 3 10 30000 10 1000 10 11 n c n m N M A f W f   + = +    = + =  0.2 0.273 0.473 1 所以强度满足要求 （2）弯矩作用平面内稳定性验算 2 0 150 200 30000 A A mm = =  = n 1 2 3 3 150 200 1000 10 6 W W mm = =   =  n 1 200 57.74 12 x i mm =  = 0 2500 43.30 91 57.74 x x x l i  = = =  2 2 1 1 0.693 1 ( / 65) 1 (43.30 / 65)   = = = + + 由构件初始偏心距 0 e = 0 ，得 k = 0 6 0 0 3 3 3 10 0.188 60 10 1 1000 10 11 1 30000 10 m c Ne M K N Wf Af +  = = =          +    +        ( ) ( ) ( ) 2 2 m = − − = − = 1 1 1 0.188 0.659 K kK 3 2 2 0 60 10 4.38 / 10 / 0.693 0.659 30000 m N N mm N mm  A  = =    弯矩作用平面内稳定性满足要求 （3）弯矩作用平面外稳定性验算 1 150 43.3 12 y i =  = 0 2500 57.74 91 43.3 y y l i  = = =  2 2 1 1 0.559 1 ( / 65) 1 (57.74/ 65) y y   = = = + + 且 2 2 4 4 2500 0.95 200 0.350 150 220 ef m m l h b k       = = =   ( ) 2 2 2 2 1 1/ 1 1/ 1 2 2 m m l m m m m c c c     +   + = − −    