k 2.形函数的性质 若l1=1;v 0 则(x,y)=N(x,y) ①.N(x2y1)=1,N(x2y)=N1(xk2yk)=0 2. N(x, y)+N,(x,y)+NK(x,y)=1 D=1 若 则(x,y )=N,+N,+N k di yi u(x,y)=a,+a2x+a,y ×C D xty D,=DD2=D2=0 由此可知:所设位移可反应单元的刚体位移.D32.形函数的性质 若 ui =1;vi = u j = vj = uk = vk = 0 u(x, y) N (x, y) 则 = i i j k =1 i u ①. Ni (xi , yi ) =1;Ni (xj , y j ) = Ni (xk , yk ) = 0 ②. Ni (x, y) + Nj (x, y) + Nk (x, y) =1 若 ui = u j = uk =1 Ni N j Nk 则 u(x, y) = + + u x y x y 1 2 3 ( , ) = + + y D D x D D D D1 2 3 = + + k k j j i i x y x y x y D 1 1 1 = k k k j j j i i i u x y u x y u x y D1 = k k j j i i u y u y u y D 1 1 1 2 = k k j j i i x u x u x u D 1 1 1 3 = D1 = D;D2 = D3 = 0 u(x, y) =1 由此可知:所设位移可反应单元的刚体位移