解:P=2A2=30×252=1875kN NIlI N △Z EA EA2 EA3 18750 0.2 4 0.2 210×10 丌D022×Q 0.0252x.00122 4 =0272mm(缩短)解: P =  2  A2 =  = 2 30 25 18.75kN l N l E A N l E A N l E A = + + 1 1 1 2 2 2 3 3 3 =   + +              18750 210 10 0 2 0 02 4 0 4 0 025 0 2 0 012 4 9 2 2 2 . . . . .   . = 0.272 mm (缩短）