如图所示的两套坐标系，一套是 Oxy ，用于实际单元，称为子单元；一套是Oξη ，它是标准后化的正方形单元，称为母单元。从母单元项子单元变换，实际上就是建立两个坐标系之间的变换关系，即: A ′ ( x , y ) = f ( A ( ξ , η ) ) A^{'}(x,y)=f(A(ξ,η)) A′(x,y)=f(A(ξ,η))

将整体坐标用插值形式表示，即： x = ∑ i = 1 N i ( ξ , η ) x i , y = ∑ i = 1 N i ( ξ , η ) y i x=\sum_{i=1}N_{i}(ξ,η)x_{i},y=\sum_{i=1}N_{i}(ξ,η)y_{i} x=i=1∑​Ni​(ξ,η)xi​,y=i=1∑​Ni​(ξ,η)yi​ ( x i , y i ) (x_{i},y_{i}) (xi​,yi​)为整体坐标系下各顶点的坐标；其中，Ni称为插值函数（形函数)。通过这一变换，对母单元上每一点(ξ,η)可以在实际单元中找到一对应点(x，y)，这样就在两个单元间建立了一一对应关系。 在实际单元中建立位移函数的插值形式： (1) u = ∑ i = 1 N i ( ξ , η ) u i , v = ∑ i = 1 N i ( ξ , η ) v i u=\sum_{i=1}N_{i}(ξ,η)u_{i},v=\sum_{i=1}N_{i}(ξ,η)v_{i}\tag{1} u=i=1∑​Ni​(ξ,η)ui​,v=i=1∑​Ni​(ξ,η)vi​(1) 因为坐标变换和位移变换都用同样个数的相应的节点值做参数，并且具有完全相同的形函数作为变换系数，所以称这种单元为等参数单元。

(2) x = ∑ i = n N i ( ξ , η ) x i ( n = i , j , k , m . . . ) , y = ∑ i = n N i ( ξ , η ) y i ( n = i , j , k , m . . . ) x=\sum_{i=n}N_{i}(ξ,η)x_{i}(n=i,j,k,m...), y=\sum_{i=n}N_{i}(ξ,η)y_{i}(n=i,j,k,m...) \tag{2} x=i=n∑​Ni​(ξ,η)xi​(n=i,j,k,m...),y=i=n∑​Ni​(ξ,η)yi​(n=i,j,k,m...)(2)

(3) { ε } = { ε x ε y γ x y } = { ∂ u ∂ x ∂ v ∂ y ∂ v ∂ x + ∂ u ∂ y } \left\{\begin{matrix} ε \end{matrix} \right\}= \left\{\begin{matrix} ε_{x}\\ ε_{y}\\ γ_{xy}\\ \end{matrix} \right\}= \left\{\begin{matrix} \frac{\partial u}{\partial x}\\ \frac{\partial v}{\partial y}\\ \frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\\ \end{matrix} \right\}\tag{3} {ε​}=⎩⎨⎧​εx​εy​γxy​​⎭⎬⎫​=⎩⎨⎧​∂x∂u​∂y∂v​∂x∂v​+∂y∂u​​⎭⎬⎫​(3) 将（1）式带入（3）式可得： { ε } = [ ∂ ∂ x 0 0 ∂ ∂ y ∂ ∂ y ∂ ∂ x ] ∗ [ N 1 N 2 N 3 N 4 . . . ] ∗ [ u 1 v 1 u 2 v 2 u 3 v 3 u 4 v 4 . . . . . . ] \left\{\begin{matrix} ε \end{matrix} \right\}= \left[ \begin{matrix} \frac{\partial }{\partial x};0\\ 0;\frac{\partial }{\partial y}\\ \frac{\partial }{\partial y};\frac{\partial }{\partial x} \end{matrix} \right]* \left[\begin{matrix} N_{1};N_{2};N_{3};N_{4};... \end{matrix} \right]* \left[\begin{matrix} u_{1};v_{1}\\ u_{2};v_{2}\\ u_{3};v_{3}\\ u_{4};v_{4}\\ ...;...\\ \end{matrix} \right] {ε​}=⎣⎡​∂x∂​0∂y∂​​0∂y∂​∂x∂​​⎦⎤​∗[N1​​N2​​N3​​N4​​...​]∗⎣⎢⎢⎢⎢⎡​u1​u2​u3​u4​...​v1​v2​v3​v4​...​⎦⎥⎥⎥⎥⎤​ 求上式 ∂ N i ∂ x ， ∂ N i ∂ y \frac{\partial N_{i}}{\partial x}，\frac{\partial N_{i}}{\partial y} ∂x∂Ni​​，∂y∂Ni​​,由复合函数求导可得： (4) ∂ N i ∂ ξ = ∂ N i ∂ x ∂ x ∂ ξ + ∂ N i ∂ y ∂ y ∂ ξ ∂ N i ∂ η = ∂ N i ∂ x ∂ x ∂ η + ∂ N i ∂ y ∂ y ∂ η \frac{\partial N_{i}}{\partial ξ}=\frac{\partial N_{i}}{\partial x}\frac{\partial x}{\partial ξ}+\frac{\partial N_{i}}{\partial y}\frac{\partial y}{\partial ξ}\newline \newline\frac{\partial N_{i}}{\partial η}=\frac{\partial N_{i}}{\partial x}\frac{\partial x}{\partial η}+\frac{\partial N_{i}}{\partial y}\frac{\partial y}{\partial η}\tag{4} ∂ξ∂Ni​​=∂x∂Ni​​∂ξ∂x​+∂y∂Ni​​∂ξ∂y​∂η∂Ni​​=∂x∂Ni​​∂η∂x​+∂y∂Ni​​∂η∂y​(4) 即: { ∂ N i ∂ ξ ∂ N i ∂ η } = [ J ] { ∂ N i ∂ x ∂ N i ∂ y } \left\{\begin{matrix} \frac{\partial N_{i}}{\partial ξ}\\ \frac{\partial N_{i}}{\partial η} \end{matrix} \right\}=[J] \left\{\begin{matrix} \frac{\partial N_{i}}{\partial x}\\ \frac{\partial N_{i}}{\partial y} \end{matrix} \right\} {∂ξ∂Ni​​∂η∂Ni​​​}=[J]{∂x∂Ni​​∂y∂Ni​​​} 其中矩阵[J]为雅可比矩阵 [ J ] = [ ∂ x ∂ ξ ∂ y ∂ ξ ∂ x ∂ η ∂ y ∂ η ] = [ ∑ ∂ N i ∂ ξ x i ∑ ∂ N i ∂ ξ y i ∑ ∂ N i ∂ η x i ∑ ∂ N i ∂ η y i ] = { ∂ N 1 ∂ ξ ∂ N 2 ∂ ξ ∂ N 3 ∂ ξ ∂ N 4 ∂ ξ ∂ N 1 ∂ η ∂ N 2 ∂ η ∂ N 3 ∂ η ∂ N 4 ∂ η } ∗ { x 1 y 1 x 2 y 2 x 3 y 3 x 4 y 4 } \left[\begin{matrix} J \end{matrix} \right]= \left[\begin{matrix} \frac{\partial x}{\partial ξ};\frac{\partial y}{\partial ξ}\\ \frac{\partial x}{\partial η};\frac{\partial y}{\partial η}\\ \end{matrix} \right]=\left[\begin{matrix} \sum\frac{\partial N_{i}}{\partial ξ}x_{i};\sum\frac{\partial N_{i}}{\partial ξ}y_{i}\\ \sum\frac{\partial N_{i}}{\partial η}x_{i};\sum\frac{\partial N_{i}}{\partial η}y_{i}\\ \end{matrix} \right]= \left\{\begin{matrix} \frac{\partial N_{1}}{\partial ξ};\frac{\partial N_{2}}{\partial ξ};\frac{\partial N_{3}}{\partial ξ};\frac{\partial N_{4}}{\partial ξ}\\ \frac{\partial N_{1}}{\partial η};\frac{\partial N_{2}}{\partial η};\frac{\partial N_{3}}{\partial η};\frac{\partial N_{4}}{\partial η}\\ \end{matrix} \right\}*\left\{\begin{matrix} x_{1};y_{1}\\ x_{2};y_{2}\\ x_{3};y_{3}\\ x_{4};y_{4}\\ \end{matrix} \right\} [J​]=[∂ξ∂x​∂η∂x​​∂ξ∂y​∂η∂y​​]=[∑∂ξ∂Ni​​xi​∑∂η∂Ni​​xi​​∑∂ξ∂Ni​​yi​∑∂η∂Ni​​yi​​]={∂ξ∂N1​​∂η∂N1​​​∂ξ∂N2​​∂η∂N2​​​∂ξ∂N3​​∂η∂N3​​​∂ξ∂N4​​∂η∂N4​​​}∗⎩⎪⎪⎨⎪⎪⎧​x1​x2​x3​x4​​y1​y2​y3​y4​​⎭⎪⎪⎬⎪⎪⎫​ 因此： { ∂ N i ∂ x ∂ N i ∂ y } = [ J ] − 1 ∗ { ∂ N i ∂ ξ ∂ N i ∂ η } \left\{\begin{matrix} \frac{\partial N_{i}}{\partial x}\\ \frac{\partial N_{i}}{\partial y} \end{matrix} \right\}=[J]^{-1}*\left\{\begin{matrix} \frac{\partial N_{i}}{\partial ξ}\\ \frac{\partial N_{i}}{\partial η} \end{matrix} \right\} {∂x∂Ni​​∂y∂Ni​​​}=[J]−1∗{∂ξ∂Ni​​∂η∂Ni​​​}

在JuliaFEM的FEMBasis包中math.jl中实现了雅可比矩阵的计算，函数需要三个参数分别为：

首先需要计算 ∂ N i ∂ ξ ， ∂ N i ∂ η \frac{\partial N_{i}}{\partial ξ}，\frac{\partial N_{i}}{\partial η} ∂ξ∂Ni​​，∂η∂Ni​​，需要用到eval_dbasis(B, xi)函数，B为单元类型，xi为待求点的局部坐标系下的坐标值。详见有限元形函数及JuliaFEM中的实现方式。

grad函数是对 { ∂ N i ∂ x ∂ N i ∂ y } \left\{\begin{matrix} \frac{\partial N_{i}}{\partial x}\\ \frac{\partial N_{i}}{\partial y} \end{matrix} \right\} {∂x∂Ni​​∂y∂Ni​​​}的求解，

由（4）式可知， { ε } = { ∂ u ∂ x ∂ v ∂ y ∂ v ∂ x + ∂ u ∂ y } \left\{\begin{matrix} ε \end{matrix} \right\}=\left\{\begin{matrix} \frac{\partial u}{\partial x}\\ \frac{\partial v}{\partial y}\\ \frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\\ \end{matrix} \right\} {ε​}=⎩⎨⎧​∂x∂u​∂y∂v​∂x∂v​+∂y∂u​​⎭⎬⎫​ u，v为各节点的整体坐标系的位移，在math.jl中另一个grad(B, T, X, xi)函数即用于求解上式的应变，此处的T为单元各节点的整体坐标系下的位移向量。求得的结果为： [ ∑ ∂ N i ∂ x u i ∑ ∂ N i ∂ x v i ∑ ∂ N i ∂ y u i ∑ ∂ N i ∂ y v i ] = [ ∂ u ∂ x ∂ v ∂ x ∂ u ∂ y ∂ v ∂ y ] \left[ \begin{matrix} \sum\frac{\partial N_{i}}{\partial x}u_{i};\sum\frac{\partial N_{i}}{\partial x}v_{i}\\ \sum\frac{\partial N_{i}}{\partial y}u_{i};\sum\frac{\partial N_{i}}{\partial y}v_{i}\\ \end{matrix} \right]= \left[ \begin{matrix} \frac{\partial u}{\partial x};\frac{\partial v}{\partial x}\\ \frac{\partial u}{\partial y};\frac{\partial v}{\partial y}\\ \end{matrix} \right] [∑∂x∂Ni​​ui​∑∂y∂Ni​​ui​​∑∂x∂Ni​​vi​∑∂y∂Ni​​vi​​]=[∂x∂u​∂y∂u​​∂x∂v​∂y∂v​​]