例2-5计算图示变截面杄的轴向变形 2-2 已知:F=15KN,l=1m,a=20mm E=200GPa N 15k 求:△l X解:作轴力图N2=F=15kN 30kN 2F=-30kN A1=A3=a2=400mm2,A2=200mm2l1=l3=0.5m,2=1m 人=A+A+A=N+N2+N2 作业: EA EA, EA 2-1(c)22-3 0.1875-0.75+0.0094=-0.844mm 2-6,2-12, 2-15例2-5 计算图示变截面杆的轴向变形 F l 2 l 2 2l 3F a a a a/2 已知：F =15kN，l = 1m，a = 20mm, E = 200GPa 求：  l 解： N1 = N2 = −2F = −30kN N3 = F =15kN N x -30 kN 15 kN 2 2 2 2 A1 = A3 = a = 400mm , A = 200mm l 1 = l 3 = 0.5m, l 2 =1m  =  +  +  = + + = 3 3 3 2 2 2 1 1 1 1 2 3 EA N l EA N l EA N l l l l l = −0.1875− 0.75+ 0.0094 = −0.844mm 作业： 2-1(c), 2-3, 2-6, 2-12， *2-15 作轴力图 1-1 2-2 3-3