②杆许可载荷: 2000 12= =692 100/12 69.3+20 02()=1.02-0.55 =0.58 100 [F]2=[12(4)A=58kN 利用变形条件 5? △l1F1l1278×3 △l,F 58× C 解得x=0.719a②杆许可载荷： [F] [ ] ( )A 58k N 0.58 100 69.3 20 ( ) 1.02 0.55 69.2 100 / 12 2000 2 2 2 2 2 = =  =      + = − = =       利用变形条件 a x F l Fl l l =   = =   58 2 27.8 3 2 2 1 1 2 1 解得 x=0.719a