Strain-displacement [B] matrix for CST

Constant strain triangle (CST)

The CST has three nodes with coordinates $(x_i,y_i)$, $i=1,2,3$ (anticlockwise) and two DOFs per node: $\{d\}=\{u_1,v_1,u_2,v_2,u_3,v_3\}^T$.

Displacement field

A linear polynomial is assumed (3 constants per component, one for each node):

$$u=\alpha_1+\alpha_2x+\alpha_3y,\qquad v=\alpha_4+\alpha_5x+\alpha_6y$$

Fitting the three nodal values of $u$ and solving for $\alpha_1,\alpha_2,\alpha_3$ gives

$$u=N_1u_1+N_2u_2+N_3u_3,\qquad N_i=\frac{1}{2A}(a_i+b_ix+c_iy)$$

where $A=\tfrac12\left|\begin{matrix}1&x_1&y_1\\1&x_2&y_2\\1&x_3&y_3\end{matrix}\right|$ is the area and

$$\begin{aligned} a_1&=x_2y_3-x_3y_2,& b_1&=y_2-y_3,& c_1&=x_3-x_2\\ a_2&=x_3y_1-x_1y_3,& b_2&=y_3-y_1,& c_2&=x_1-x_3\\ a_3&=x_1y_2-x_2y_1,& b_3&=y_1-y_2,& c_3&=x_2-x_1 \end{aligned}$$

The same $N_i$ are used for $v$. In matrix form $\{u\ v\}^T=[N]\{d\}$ with $[N]=\begin{bmatrix}N_1&0&N_2&0&N_3&0\\0&N_1&0&N_2&0&N_3\end{bmatrix}$.

Strain-displacement relation

For plane problems

$$\{\varepsilon\}=\begin{Bmatrix}\varepsilon_x\\\varepsilon_y\\\gamma_{xy}\end{Bmatrix}=\begin{Bmatrix}\partial u/\partial x\\\partial v/\partial y\\\partial u/\partial y+\partial v/\partial x\end{Bmatrix}$$

Differentiating, $\dfrac{\partial N_i}{\partial x}=\dfrac{b_i}{2A}$ and $\dfrac{\partial N_i}{\partial y}=\dfrac{c_i}{2A}$ (the constants $a_i$ drop out). So

$$\varepsilon_x=\frac{1}{2A}(b_1u_1+b_2u_2+b_3u_3),\quad \varepsilon_y=\frac{1}{2A}(c_1v_1+c_2v_2+c_3v_3),$$ $$\gamma_{xy}=\frac{1}{2A}(c_1u_1+b_1v_1+c_2u_2+b_2v_2+c_3u_3+b_3v_3)$$

Hence $\{\varepsilon\}=[B]\{d\}$ with

$$\boxed{[B]=\frac{1}{2A}\begin{bmatrix}b_1&0&b_2&0&b_3&0\\0&c_1&0&c_2&0&c_3\\c_1&b_1&c_2&b_2&c_3&b_3\end{bmatrix}}$$

Remarks

• $[B]$ contains only the nodal coordinates, so it is constant over the element. Strain and stress (from $\{\sigma\}=[D][B]\{d\}$) are therefore constant, hence the name "constant strain" triangle.
• The element stiffness follows as $[k]=t\,A\,[B]^T[D][B]$ for an element of thickness $t$.