c)应用位移边界条件和连续条件求积分常数 x1=0,w1=0;x2=l,w2=0 x1=x2=a,w1=w2;w'1=w2 o(12-b2);D1=D2=0 d)确定挠曲线和转角方程 W 12-b leL Fb B=w (2-b2)-3x 6El Fb (x2-a)3-x2+(2-b2) 6EⅠb Fb X -a (72-b 2/EI
d) 确定挠曲线和转角方程 c) 应用位移边界条件和连续条件求积分常数 x1 = 0 , w1 = 0 ; x2 = l , w2 = 0 . x1= x2 = a ,w1 = w2 ;w'1 = w'2 2 2 1 2 1 2 ( ); 0 6 Fb C C l b D D l = = − = = 1 2 2 2 1 1 2 2 2 1 1 1 6 ( ) 3 6 z z Fbx w l b x lEI Fb w l b x lEI = − − = = − − 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ( ) ( ) 6 1 ( ) ( ) 2 3 z z Fb l w x a x l b x lEI b Fb l w x a x l b lEI b = − − + − = = − − + −
Fbx Fb b (x2-a)3-x2+(-b)x2 lEI leL. b Fb F6 6/B(12-b2)-3x 2lE(x2-a)-x2+2(-b) e)跨中点挠度及两端端截面的转角 F6 Fb vx548(312-4b2),0=v1--24E 12-4b 两端支座处的转角 o Fab(l+b Fab(+a) 6/EI 6/EI
e) 跨中点挠度及两端端截面的转角 2 2 2 2 2 2 (3 4 ); 4 48 24 l l x x z z Fb Fb w l b w l b EI lEI = = = − = = − ( ) ( ) ; 6 6 A B z z Fab l b Fab l a lEI lEI + + = = − 两端支座处的转角—— 1 2 2 2 1 1 2 2 2 1 1 1 6 ( ) 3 6 z z Fbx w l b x lEI Fb w l b x lEI = − − = = − − 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ( ) ( ) 6 1 ( ) ( ) 2 3 z z Fb l w x a x l b x lEI b Fb l w x a x l b lEI b = − − + − = = − − + −