δ11=ΣFN12EA=(0.8X4X2+0.6×3×2+1X5×2)yEA=17.28/EA △IP=ΣFNI FNPUEA =0.8×13.33×4-0.8×6.67×4+0.6×5×3+1×8.33×5)EA=71.96/EA 3)代入力法方程中，求解x1 x1=-△1P611=-4.17KN 4)叠加计算各杆轴力 FN-FNIX1+FNP X1=1 r-0.d 06-0.8D0 F图 D B FN图 7-11.试绘出图示连续梁的M图，并作校核。已知I=36×10m,E=3×10'kPa h209 4m 4m 4m 4m4m4m 基本体系11 = ∑FN1 2 l/EA= (0.82×4×2+0.62×3×2+12×5×2)/EA=17.28/EA 1P = ∑FN1 FNPl/EA =(0.8×13.33×4-0.8×6.67×4+0.6×5×3+1×8.33×5)/EA=71.96/EA 3)代入力法方程中，求解 x1 x1 = — 1P /11 = —4.17KN 4) 叠加计算各杆轴力 FN =FN1x1 +FNP FN1 图 F N 图 7-11.试绘出图示连续梁的 M 图，并作校核。已知 I=36×10-4 m 4 ,E=3×107 kPa 基本体系