§15基于变形体虚位移原理的弯曲单元(自由式单元的单元分析 、确定形函数 f(1)+(1-1)f(1),=0 2、试凑法 dx 利用形函数的性质建立形函数矩阵f()=(1-5)g() (1)确定N1() N2=(1-5)2g() 由5=1,M=0可设 dw (1-5)g(2)+(1-)2g(2) (1-5)f(2) dx 由5=0,N1=1可知 dx =-1(-0g(0+0-0)2g(0,7=0 f()=1 2 所以 g(0)+g(0)·=0 (2)确定N2(5) N2(O)=g(0)=1 由5=1,N2=0可设 g(0)=2g(5)=1+25 N2=(1-f() N2(5)=(1-5)(1+25)=1-352+25 a5=0dN2l0=0 dx N(0)=1M(l)=0N4(0)=0N()=1 N20)=1M2)=0N0)=0N()=1 2=-f(5)+(1-5)f dx§1.5 基于变形体虚位移原理的弯曲单元(自由式单元)的单元分析 一、确定形函数 2、试凑法 利用形函数的性质建立形函数矩阵 (1)确定 ( ) 1 N  1, 0 由  = N1 = 可设 (1 ) ( ) N1 = − f  由  = 0,N1 =1 可知 f ( ) =1 所以 N1 =1− (2)确定 ( ) 2 N  由  =1,N2 = 0 可设 (1 ) ( ) N2 = − f  0; 0 0 2 1 2  = =  = = dx dN dx dN l f f dx l dN 1 ( ) (1 ) 1 2 = −  + −  0 1 (1) (1 1) (1) 1 1 2 = = − + −  = l f f dx l dN  f ( ) = (1− )g( ) (1 ) ( ) 2 N2 = − g  l g g dx l dN 1 (1 ) ( ) (1 ) ( ) 2 2 2 = − −  + −    (0) 0 (1) 0 (0) 0 (1) 0 (0) 1 (1) 0 (0) 0 (1) 1 (0) 1 (1) 0 (0) 0 (1) 1 3 3 6 6 2 2 5 5 1 1 4 4 = = = = = = = = = = = = N N N N N N N N N N N N 0 1 (1 0) (0) (1 0) (0) 2 2 0 2 = = − − + −   = l g g dx l dN  0 1 (0) (0) 2 − +   = l g g l (0) (0) 1 N2 = g = g (0) = 2 g( ) =1+ 2 2 2 3 2 N () = (1−) (1+ 2) =1−3 + 2