Question p.6.13: A constant strain triangular element has corners 1(0,0), 2(4...
A constant strain triangular element has corners 1(0,0), 2(4,0) and 3(2,2) and is 1 unit thick. If the elasticity matrix [D] has elements D_{11}=D_{22}=a, D_{12}=D_1=b, \mathrm{D}_{13}=D_{23}=D_{31}=D_{32}=0 \text { and } \mathrm{D}_{33}=c derive the stiffness matrix for the element.
Assume displacement functions
\begin{aligned}&u(x, y)=\alpha_1+\alpha_2 x+\alpha_3 y \\&v(x, y)=\alpha_4+\alpha_5 x+\alpha_6 y\end{aligned}Then
\begin{aligned}&u_1=\alpha_1 \\&u_2=\alpha_1+4 \alpha_2 \\&u_3=\alpha_1+2 \alpha_2+2 \alpha_3\end{aligned}Solving
\alpha_2=\frac{u_2-u_1}{4} \quad \alpha_3=\frac{2 u_3-u_1-u_2}{4}Therefore
u=u_1+\left(\frac{u_2-u_1}{4}\right) x+\left(\frac{2 u_3-u_1-u_2}{4}\right) yor
u=\left(1-\frac{x}{4}-\frac{y}{4}\right) u_1+\left(\frac{x}{4}-\frac{y}{4}\right) u_2+\frac{y}{2} u_3Similarly
v=\left(1-\frac{x}{4}-\frac{y}{4}\right) v_1+\left(\frac{x}{4}-\frac{y}{4}\right) v_2+\frac{y}{2} v_3Then, from Eqs (1.18) and (1.20)
\left.\begin{array}{rl}\varepsilon_x & =\frac{\partial u}{\partial x} \\\varepsilon_y & =\frac{\partial v}{\partial y} \\\varepsilon_z & =\frac{\partial w}{\partial z}\end{array}\right\} (1.18)
\left.\begin{array}{l}\gamma_{x z}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z} \\\gamma_{x y}=\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y} \\\gamma_{y z}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}\end{array}\right\} (1.20)
\begin{aligned}\varepsilon_x &=\frac{\partial u}{\partial x}=-\frac{u_1}{4}+\frac{u_2}{4} \\\varepsilon_y &=\frac{\partial v}{\partial y}=-\frac{v_1}{4}-\frac{v_2}{4}+\frac{v_3}{2} \\\gamma_{x y} &=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=-\frac{u_1}{4}-\frac{u_2}{4}-\frac{v_1}{4}+\frac{v_2}{4}\end{aligned}Hence
[B]\left\{\delta^e\right\}=\left[\begin{array}{c}\frac{\partial u}{\partial x} \\\frac{\partial v}{\partial y} \\\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\end{array}\right]=\frac{1}{4}\left[\begin{array}{rrrrrc}-1 & 0 & 1 & 0 & 0 & 0 \\0 & -1 & 0 & -1 & 0 & 2 \\-1 & -1 & -1 & 1 & 2 & 0\end{array}\right]\left\{\begin{array}{l}u_1 \\v_1 \\u_2 \\v_2 \\u_3 \\v_3\end{array}\right\}But
[D]=\left[\begin{array}{lll}a & b & 0 \\b & a & 0 \\0 & 0 & c\end{array}\right]so that
[D][B]=\frac{1}{4}\left[\begin{array}{rrrrrr}-a & -b & a & -b & 0 & 2 b \\-b & -a & b & -a & 0 & 2 a \\-c & -c & -c & c & 2 c & 0\end{array}\right]and
\text { Since }\left[K^e\right]=[B]^{\mathrm{T}}[D][B] \times 4 \times 1