CST equilateral plate: stresses at centroid

Data: plane stress, one CST element, side $=100$ mm, $t=10$ mm, $E=210\times10^3$ MPa, $\nu=0.3$, unit weight $78.5\times10^{-6}$ N/mm$^3$. Nodes: 1 $(0,0)$, 2 $(100,0)$, 3 $(50,\ 86.603)$. Units: N, mm.

Step 1: Geometry

• Node 1: $(0,\ 0)$ mm
• Node 2: $(100,\ 0)$ mm
• Node 3: $(50,\ 86.6025)$ mm

Area: $A=\tfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=4330.13\ \text{mm}^2$

$b_1=y_2-y_3=-86.6025,\ b_2=y_3-y_1=86.6025,\ b_3=y_1-y_2=0$ mm

$c_1=x_3-x_2=-50,\ c_2=x_1-x_3=-50,\ c_3=x_2-x_1=100$ mm

Step 2: [B] and [D] matrices (plane stress)

$$[B]=\frac{1}{2A}\begin{bmatrix}-86.60 & 0 & 86.60 & 0 & 0 & 0 \\ 0 & -50 & 0 & -50 & 0 & 100 \\ -50 & -86.60 & -50 & 86.60 & 100 & 0\end{bmatrix},\quad 2A=8660.25\ \text{mm}^2$$ $$[D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix}=10^{5}\begin{bmatrix}2.308 & 0.692 & 0 \\ 0.692 & 2.308 & 0 \\ 0 & 0 & 0.808\end{bmatrix}\ \text{N/mm}^2$$

Step 3: Element stiffness matrix

$$[k]=t\,A\,[B]^T[D][B]= 10^{6}\begin{bmatrix}1.116 & 0.375 & -0.883 & -0.029 & -0.233 & -0.346 \\ 0.375 & 0.683 & 0.029 & -0.017 & -0.404 & -0.666 \\ -0.883 & 0.029 & 1.116 & -0.375 & -0.233 & 0.346 \\ -0.029 & -0.017 & -0.375 & 0.683 & 0.404 & -0.666 \\ -0.233 & -0.404 & -0.233 & 0.404 & 0.466 & 0 \\ -0.346 & -0.666 & 0.346 & -0.666 & 0 & 1.332\end{bmatrix}\ \text{N/mm}$$

(DOF order: $u_1,v_1,u_2,v_2,u_3,v_3$.)

Step 4: Load vector

The 10 kN/m on the inclined edge 2-3 (length 100 mm) gives a total of $10\times0.1=1$ kN $=1000$ N. Assumed to act vertically downward, it is shared equally: 500 N at node 2 (absorbed by the support) and 500 N at node 3. Self-weight $W=78.5\times10^{-6}\times10\times4330.1=3.399$ N, $1.133$ N downward at each node. Load on the free node: $F_{y3}=-501.13$ N.

$$\{F\}=\begin{Bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ -501.1\end{Bmatrix}\ \text{N}$$

Boundary conditions and solution

Nodes 1 and 2 are on the supported base ($u=v=0$). Free DOFs: $u_3,v_3$ (rows/columns 5 and 6).

$$10^{6}\begin{bmatrix}0.466 & 0 \\ 0 & 1.332\end{bmatrix}\begin{Bmatrix}u_3 \\ v_3\end{Bmatrix}=\begin{Bmatrix}0 \\ -501.1\end{Bmatrix}$$ $$\begin{aligned} u_3&=0\ \text{mm} \\ v_3&=-3.761\times10^{-4}\ \text{mm} \end{aligned}$$

Strains and stresses (constant over the element, so equal at the centroid)

Centroid: $(50.00,\ 28.87)$ mm.

$$\{\varepsilon\}=[B]\{d\}=\begin{Bmatrix}0 \\ -4.343\times10^{-6} \\ 0\end{Bmatrix}$$ $$\{\sigma\}=[D]\{\varepsilon\}=\begin{Bmatrix}-0.3007 \\ -1.002 \\ 0\end{Bmatrix}\ \text{MPa}$$

Answer: $u_3=0$ mm, $v_3=-3.76\times10^{-4}$ mm. At the centroid $\varepsilon_x=0$, $\varepsilon_y=-4.34\times10^{-6}$, $\gamma_{xy}=0$; $\sigma_x=-0.301$, $\sigma_y=-1.002$, $\tau_{xy}=0.000$ N/mm$^2$ (compressive).