CST plate: stiffness, load vector, displacements

Assumptions: plane stress, one CST element. Nodes: 1 $(0,0)$ pinned, 2 $(800,0)$ roller, 3 $(0,600)$. Given $E=2\times10^5$ MPa and $G=105\times10^3$ MPa, the Poisson's ratio follows from $G=\dfrac{E}{2(1+\nu)}$:

$$\nu=\frac{E}{2G}-1=\frac{2\times10^5}{2\times105\times10^3}-1=-0.0476$$

Units: N, mm.

Step 1: Geometry

• Node 1: $(0,\ 0)$ mm
• Node 2: $(800,\ 0)$ mm
• Node 3: $(0,\ 600)$ mm

Area: $A=\tfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=240000\ \text{mm}^2$

$b_1=y_2-y_3=-600,\ b_2=y_3-y_1=600,\ b_3=y_1-y_2=0$ mm

$c_1=x_3-x_2=-800,\ c_2=x_1-x_3=0,\ c_3=x_2-x_1=800$ mm

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

$$[B]=\frac{1}{2A}\begin{bmatrix}-600 & 0 & 600 & 0 & 0 & 0 \\ 0 & -800 & 0 & 0 & 0 & 800 \\ -800 & -600 & 0 & 600 & 800 & 0\end{bmatrix},\quad 2A=480000\ \text{mm}^2$$ $$[D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix}=10^{5}\begin{bmatrix}2.005 & -0.095 & 0 \\ -0.095 & 2.005 & 0 \\ 0 & 0 & 1.050\end{bmatrix}\ \text{N/mm}^2$$

Step 3: Element stiffness matrix

$$[k]=t\,A\,[B]^T[D][B]= 10^{6}\begin{bmatrix}1.452 & 0.477 & -0.752 & -0.525 & -0.700 & 0.048 \\ 0.477 & 1.730 & 0.048 & -0.394 & -0.525 & -1.336 \\ -0.752 & 0.048 & 0.752 & 0 & 0 & -0.048 \\ -0.525 & -0.394 & 0 & 0.394 & 0.525 & 0 \\ -0.700 & -0.525 & 0 & 0.525 & 0.700 & 0 \\ 0.048 & -1.336 & -0.048 & 0 & 0 & 1.336\end{bmatrix}\ \text{N/mm}$$

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

Step 4: Load vector

Self-weight $W=\gamma tA=78.5\times10^{-6}\times10\times240000=188.4$ N, i.e. $62.8$ N downward at each node. The 5 kN/m$^2$ = 0.005 N/mm$^2$ on edge 1-3 (600 mm long, acting horizontally toward $+x$, assumed) gives $0.005\times10\times600=30$ N, 15 N to each end node. The full nodal load vector is

$$\{F\}=\begin{Bmatrix}F_{u1}\\F_{v1}\\F_{u2}\\F_{v2}\\F_{u3}\\F_{v3}\end{Bmatrix}=\begin{Bmatrix}15 \\ -62.8 \\ 0 \\ -62.8 \\ 15 \\ -62.8\end{Bmatrix}\ \text{N}$$

The loads at the restrained DOFs ($u_1$, $v_1$, $v_2$) go straight into the supports, so the loads on the free DOFs are $F_{u2}=0$, $F_{u3}=15$ N and $F_{v3}=-62.8$ N.

Boundary conditions and solution

Node 1 is pinned ($u_1=v_1=0$) and node 2 is a roller ($v_2=0$, $u_2$ free). Unknowns: $u_2,u_3,v_3$ (rows/columns 3, 5, 6).

$$10^{6}\begin{bmatrix}0.752 & 0 & -0.048 \\ 0 & 0.700 & 0 \\ -0.048 & 0 & 1.336\end{bmatrix}\begin{Bmatrix}u_2 \\ u_3 \\ v_3\end{Bmatrix}=\begin{Bmatrix}0 \\ 15 \\ -62.8\end{Bmatrix}$$ $$\begin{aligned} u_2&=-2.99\times10^{-6}\ \text{mm} \\ u_3&=2.143\times10^{-5}\ \text{mm} \\ v_3&=-4.71\times10^{-5}\ \text{mm} \end{aligned}$$

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

Centroid: $(266.67,\ 200.00)$ mm.

$$\{\varepsilon\}=[B]\{d\}=\begin{Bmatrix}-3.738\times10^{-9} \\ -7.85\times10^{-8} \\ 3.571\times10^{-8}\end{Bmatrix}$$ $$\{\sigma\}=[D]\{\varepsilon\}=\begin{Bmatrix}0 \\ -0.0157 \\ 0.00375\end{Bmatrix}\ \text{MPa}$$

Answer: $[k]$ and $\{F\}$ as above; $u_2=-2.99\times10^{-6}$ mm, $u_3=2.14\times10^{-5}$ mm, $v_3=-4.71\times10^{-5}$ mm, with $u_1=v_1=v_2=0$.