设截面A形心为座标原点,X轴正向向右、Y轴正向向下
弯矩方程 M(x) = -F(L/2 -x) -FL = F.x -(3/2)FL
AC段挠曲线近似二阶微分方程 EI.y"ac = -M(X) = (3/2)FL -F.x
一次积分得：EI.y'ac = (3/4)FL.x -(F/2).(x^2) +D1 , ①
再次积分得: EI.yac = (3/8)FL.X^2 -(F/6).(X^3) +D1.x +D2 , ②
A处边界条件：x=0, y'=0, y =0, 代入①②式得：D1=0, D2=0
即： θac =y'ac = [(3/4)FL.x -(F/2).(x^2)]/EI ,③
yac =[(3/8)FL.X^2 -(F/6).(X^3)]/EI ,④
将x = L/2 代入③、④式,得截面C的转角及挠度：
θc =[(1/4)F.L^2]/EI
yc =[(1/96)F.L^3]/EI
梁自由端转角θmax =θc =[(1/4)F.L^2]/EI
按叠加原理计算梁自由端的最大挠度:
ymax = yc +(L/2).θc
=[(1/96)F.L^3]/EI +(L/2)[(1/4)F.L^2]/EI
=[(13/96）F.L^3]/EI