应用位移边界条件求积分常数 EJ(0)=:Pa+C2=0 6 E/6(0=-Pa2+C1=0 2 f (a)=6(a):C1=D1 f(a=f(a) Ca+C2=Da+D C=D2aC2≈D1 Pa 6应用位移边界条件求积分常数 0 6 1 (0) 2 3 EIf = Pa +C = 0 2 1 (0) 1 2 EI = − Pa +C = 3 2 2 2 1 1 6 1 ; 2 1 C = D = Pa C = D = − Pa ( ) ( ) − + f a = f a ( ) ( ) − +  a = a C1 = D1 C1 a +C2 = D1 a + D2 P L a x f