Question 9.5: Derive the stiffness matrix in global coordinates for a thre...

With reference to §9.9 and with respect to the node labelling shown in Fig. 9.41, matrix [A] will be given as:

[A]=\begin{bmatrix} 1 & x_{1} & y_{1} & 0 & 0 & 0 \\ 1 & x_{2} & y_{2} & 0 & 0 & 0 \\1 & x_{3} & y_{3} & 0 & 0 & 0\\0 & 0 & 0 & 1 & x_{1} & y_{1} \\0 & 0 & 0 & 1 & x_{2} & y_{2} \\0 & 0 & 0 & 1 & x_{3} & y_{3}\end{bmatrix} =\begin{bmatrix} A_{\alpha \alpha } & A_{\alpha \beta } \\ A_{\beta \alpha } & A_{\beta \alpha } \end{bmatrix}

Then [A]^{-1}=\begin{bmatrix} [A_{\alpha \alpha }]^{-1} & 0 \\ 0 & [A_{\beta \beta }]^{-1} \end{bmatrix}   where  [A_{\alpha \alpha }]^{-1}= [A_{\beta \beta }]^{-1}  in this case

Obtaining the inverse of the partition

adj [A_{\alpha \alpha }]=[C_{\alpha \alpha }]^{T}=\begin{bmatrix} x_{2}y_{3}-x_{3}y_{2} & y_{2}-y_{3} & x_{3}-x_{2}\\ x_{3}y_{1}-x_{1}y_{3} & y_{3}-y_{1} & x_{1}-x_{3} \\x_{1}y_{2}-x_{2}y_{1} & y_{1}-y_{2} & x_{2}-x_{1}\end{bmatrix} ^{T}

=\begin{bmatrix} x_{2}y_{3}-x_{3}y_{2} & x_{3}y_{1}-x_{1}y_{3} & x_{1}y_{2}-x_{2}y_{1} \\ y_{2}-y_{3} & y_{3}-y_{1} &y_{1}-y_{2}\\x_{3}-x_{2} & x_{1}-x_{3} & x_{2}-x_{1}\end{bmatrix}

 and det [A_{\alpha \alpha }]=( x_{2}y_{3}-x_{3}y_{2})-(x_{1}y_{3}-x_{3}y_{1})+(x_{1}y_{2}-x_{2}y_{1})
=x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})
=2 \times  area of element = 2a ,(see following derivation )

Then [A_{\alpha \alpha }]^{-1}=\frac{adj [A_{\alpha \alpha }]}{det [A_{\alpha \alpha }]}=\frac{1}{2a} \begin{bmatrix} x_{2}y_{3}-x_{3}y_{2} & x_{3}y_{1}-x_{1}y_{3} & x_{1}y_{2}-x_{2}y_{1} \\ y_{2}-y_{3} & y_{3}-y_{1} &y_{1}-y_{2}\\x_{3}-x_{2} & x_{1}-x_{3} & x_{2}-x_{1}\end{bmatrix}

Hence,  [A]^{-1}=\frac{1}{2a} \begin{bmatrix} x_{2}y_{3}-x_{3}y_{2} & x_{3}y_{1}-x_{1}y_{3} & x_{1}y_{2}-x_{2}y_{1} &0&0&0\\ y_{2}-y_{3} & y_{3}-y_{1} &y_{1}-y_{2}&0&0&0\\x_{3}-x_{2} & x_{1}-x_{3} & x_{2}-x_{1}&0&0&0 \\ 0&0&0&x_{2}y_{3}-x_{3}y_{2} & x_{3}y_{1}-x_{1}y_{3} & x_{1}y_{2}-x_{2}y_{1} \\0&0&0& y_{2}-y_{3} & y_{3}-y_{1} &y_{1}-y_{2}\\ 0&0&0&x_{3}-x_{2} & x_{1}-x_{3} & x_{2}-x_{1}\end{bmatrix}

Area of element

With reference to Fig 9.42, area of triangular element ,

a = area of enclosing rectangle – (area of triangles b,c and d)

=(x_{2}-x_{1})(y_{3}-y_{1})-(1/2)(x_{2}-x_{1})(y_{2}-y_{1})-(1/2)(x_{2}-x_{3})(y_{3}-y_{2})-(1/2)(x_{3}-x_{1})(y_{3}-y_{1})
=x_{2}y_{3}-x_{2}y_{1}-x_{1}y_{3}+x_{1}y_{1}-(1/2)[x_{2}y_{2}-x_{2}y_{1}-x_{1}y_{2}+x_{1}y_{1})+(x_{2}y_{3}-x_{2}y_{2}-x_{3}y_{3}+x_{3}y_{2})+(x_{3}y_{3}-x_{3}y_{1}-x_{1}y_{3}+x_{1}y_{1})]
=(1/2)(x_{2}y_{3}-x_{2}y_{1}-x_{1}y_{3}+x_{1}y_{2}-x_{3}y_{2}+x_{3}y_{1})
=(1/2)[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]

§9.9 gives matrix [B] as

[B]=\begin{bmatrix} 0 & 1 &0&0&0&0 \\ 0 & 0 &0&0&0&1 \\0 & 0 &1&0&1&0 \end{bmatrix} [A]^{-1}

Substituting for [A]^{-1} from above and evaluating the product gives

The required element stiffness matrix can now be found by substituting into the relatio

[k] = at [B]^{T}[D][B]

where the abbreviation y_{23} denotes y_{2} – y_{3}, etc.
Choosing to evaluate the product [D][B] first, gives

Completing the matrix multiplication, reversing the sequence of some of the coordinates so that all subscripts are in descending order, gives the required element stiffness matrix as