Derive CST element stiffness matrix

Displacement and strain

The element has nodes 1, 2, 3 with $\{d\} = \{u_1, v_1, u_2, v_2, u_3, v_3\}^T$. The displacement field is $\{u\} = [N]\{d\}$ with linear shape functions $N_i = \dfrac{1}{2A}(a_i + b_ix + c_iy)$.

Strains:

$$\{\varepsilon\} = \begin{Bmatrix} \partial u/\partial x \\ \partial v/\partial y \\ \partial u/\partial y + \partial v/\partial x \end{Bmatrix} = [B]\{d\},\quad [B] = \frac{1}{2A}\begin{bmatrix} b_1 & 0 & b_2 & 0 & b_3 & 0 \\ 0 & c_1 & 0 & c_2 & 0 & c_3 \\ c_1 & b_1 & c_2 & b_2 & c_3 & b_3 \end{bmatrix}$$

Stresses: $\{\sigma\} = [D]\{\varepsilon\} = [D][B]\{d\}$.

Potential energy

$$\Pi = \underbrace{\frac{1}{2}\int_V \{\varepsilon\}^T[D]\{\varepsilon\}\,dV}_{U} - \underbrace{\left(\int_V \{u\}^T\{X\}\,dV + \int_S \{u\}^T\{T\}\,dS + \{d\}^T\{P\}\right)}_{W}$$

Substituting $\{\varepsilon\} = [B]\{d\}$:

$$\Pi = \frac{1}{2}\{d\}^T\left[\int_V [B]^T[D][B]\,dV\right]\{d\} - \{d\}^T\{f\}$$

Minimising, $\dfrac{\partial\Pi}{\partial\{d\}} = 0$, gives $[k]\{d\} = \{f\}$ with

$$[k] = \int_V [B]^T[D][B]\,dV$$

Evaluation

For CST, $[B]$ and $[D]$ are constant, and the integral over the volume is just the element volume $tA$ ($t$ is the thickness):

$$\boxed{[k] = t\,A\,[B]^T[D][B]}$$

This is a symmetric $6\times6$ matrix, singular until supports are applied (three rigid-body modes: two translations and one rotation).

Load vectors

• Body force $\{X\} = \{X_x, X_y\}^T$ (per unit volume): $\{f_b\} = \displaystyle\int_V [N]^T\{X\}\,dV = \dfrac{tA}{3}\{X_x,\ X_y,\ X_x,\ X_y,\ X_x,\ X_y\}^T$, i.e. one-third of the total body force at each node.
• Uniform traction $\{T\} = \{T_x, T_y\}^T$ on an edge of length $\ell$ between nodes $i$ and $j$: $\{f_s\} = \displaystyle\int_S [N]^T\{T\}\,t\,dS$ gives $\dfrac{t\,\ell}{2}T_x$ and $\dfrac{t\,\ell}{2}T_y$ at each of the two edge nodes and zero at the third node.