CST plate: deflection and stresses at centroid

A constant strain triangle (CST) is used with plane stress ($E=69900$ MPa, $\nu=0.3$, $t=8$ mm). Units: N, mm. Node 1 is the loaded node; the nodes 2 and 3 on the left edge are fixed.

Step 1: Geometry

• Node 1: $(35,\ 25)$ mm
• Node 2: $(0,\ 25)$ mm
• Node 3: $(0,\ 0)$ mm

Area: $A=\tfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=437.5\ \text{mm}^2$

$b_1=y_2-y_3=25,\ b_2=y_3-y_1=-25,\ b_3=y_1-y_2=0$ mm

$c_1=x_3-x_2=0,\ c_2=x_1-x_3=35,\ c_3=x_2-x_1=-35$ mm

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

$$[B]=\frac{1}{2A}\begin{bmatrix}25 & 0 & -25 & 0 & 0 & 0 \\ 0 & 0 & 0 & 35 & 0 & -35 \\ 0 & 25 & 35 & -25 & -35 & 0\end{bmatrix},\quad 2A=875\ \text{mm}^2$$ $$[D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix}=10^{4}\begin{bmatrix}7.681 & 2.304 & 0 \\ 2.304 & 7.681 & 0 \\ 0 & 0 & 2.688\end{bmatrix}\ \text{N/mm}^2$$

Step 3: Element stiffness matrix

$$[k]=t\,A\,[B]^T[D][B]= 10^{5}\begin{bmatrix}2.195 & 0 & -2.195 & 0.922 & 0 & -0.922 \\ 0 & 0.768 & 1.075 & -0.768 & -1.075 & 0 \\ -2.195 & 1.075 & 3.700 & -1.997 & -1.506 & 0.922 \\ 0.922 & -0.768 & -1.997 & 5.070 & 1.075 & -4.302 \\ 0 & -1.075 & -1.506 & 1.075 & 1.506 & 0 \\ -0.922 & 0 & 0.922 & -4.302 & 0 & 4.302\end{bmatrix}\ \text{N/mm}$$

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

Step 4: Load vector

Loads act at node 1 only. Assumed directions: 45 N horizontal toward $+x$ and 90 N vertical downward ($-y$). Self-weight is neglected (not given).

$$\{F\}=\begin{Bmatrix}45 \\ -90 \\ 0 \\ 0 \\ 0 \\ 0\end{Bmatrix}\ \text{N}$$

Boundary conditions and solution

The left edge is fixed, so nodes 2 and 3 have $u=v=0$. Only $u_1$ and $v_1$ are unknown, so use rows and columns 1 and 2 of $[k]$.

$$10^{5}\begin{bmatrix}2.195 & 0 \\ 0 & 0.768\end{bmatrix}\begin{Bmatrix}u_1 \\ v_1\end{Bmatrix}=\begin{Bmatrix}45 \\ -90\end{Bmatrix}$$ $$\begin{aligned} u_1&=2.05\times10^{-4}\ \text{mm} \\ v_1&=-0.001172\ \text{mm} \end{aligned}$$

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

Centroid: $(11.67,\ 16.67)$ mm.

$$\{\varepsilon\}=[B]\{d\}=\begin{Bmatrix}5.858\times10^{-6} \\ 0 \\ -3.348\times10^{-5}\end{Bmatrix}$$ $$\{\sigma\}=[D]\{\varepsilon\}=\begin{Bmatrix}0.45 \\ 0.135 \\ -0.9\end{Bmatrix}\ \text{MPa}$$

Answer: deflection at the load point (node 1): $u_1=2.05\times10^{-4}$ mm (horizontal), $v_1=-0.00117$ mm (vertical, downward). Stresses at the centroid: $\sigma_x=0.450$, $\sigma_y=0.135$, $\tau_{xy}=-0.900$ N/mm$^2$.