CST: nodal deformations and stresses

The plate is idealised as one constant strain triangle (CST) in plane stress. Units: kN and m.

Assumptions

• Nodes: 1 (0, 0), 2 (0.8, 0), 3 (0, 0.6) m, numbered anticlockwise. Thickness $t = 0.01$ m.
• The base nodes 1 and 2 are fixed ($u_1=v_1=u_2=v_2=0$); node 3 is free. The figure gives no other support, so this is the usual reading.
• $E = 2\times10^{8}$ kN/m$^2$ and $G = 105\times10^{6}$ kN/m$^2$ are used exactly as given, so $\nu = \dfrac{E}{2G}-1 = -0.0476$ (the given $G$ is not consistent with $\nu=0.3$; the stated data are followed).

Step 1: Geometry

$$A = \tfrac12(0.8)(0.6) = 0.24\ \text{m}^2$$

With $b_1=y_2-y_3=-0.6,\ b_2=y_3-y_1=0.6,\ b_3=y_1-y_2=0$ and $c_1=x_3-x_2=-0.8,\ c_2=x_1-x_3=0,\ c_3=x_2-x_1=0.8$:

$$[B]=\frac{1}{2A}\begin{bmatrix} -0.6 & 0 & 0.6 & 0 & 0 & 0 \\ 0 & -0.8 & 0 & 0 & 0 & 0.8 \\ -0.8 & -0.6 & 0 & 0.6 & 0.8 & 0 \end{bmatrix}$$

Step 2: Elasticity matrix (plane stress)

$$[D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\ 0&0&\frac{1-\nu}{2}\end{bmatrix} = \begin{bmatrix} 2.0045\times10^{8} & -9545455 & 0 \\ -9545455 & 2.0045\times10^{8} & 0 \\ 0 & 0 & 1.05\times10^{8} \end{bmatrix}\ \text{kN/m}^2$$

(the shear term is $G$).

Step 3: Element stiffness $[k]=tA[B]^T[D][B]$

Only the rows and columns of the free dofs $(u_3, v_3)$ are needed:

$$[k]_{33}=\begin{bmatrix} 700000 & 0 \\ 0 & 1336364 \end{bmatrix}\ \text{kN/m}$$

Step 4: Nodal loads

• Edge load: $50\times0.01\times0.6 = 0.30$ kN in the $-x$ direction, shared equally by nodes 1 and 3, so $F_{3x}=-0.15$ kN.
• Self weight: $W = 78.5\times0.24\times0.01 = 0.1884$ kN, one third to each node, so $F_{3y}=-0.0628$ kN.

Step 5: Nodal deformations

$$\begin{bmatrix}700000 & 0\\ 0 & 1336363.6\end{bmatrix}\begin{bmatrix}u_3\\ v_3\end{bmatrix}=\begin{bmatrix}-0.15\\ -0.0628\end{bmatrix}$$ $$u_3 = -2.143\times10^{-7}\ \text{m},\qquad v_3 = -4.699\times10^{-8}\ \text{m}$$

All other nodal displacements are zero.

Step 6: Stresses, $\{\sigma\}=[D][B]\{u\}$

$$\{\sigma\}=\begin{bmatrix}\sigma_x\\ \sigma_y\\ \tau_{xy}\end{bmatrix}=\begin{bmatrix}0.748\\ -15.70\\ -37.50\end{bmatrix}\ \text{kN/m}^2$$

Check: the element force at node 3 in the $x$ direction, from $tA[B]^T\{\sigma\}$, is $\tfrac{t}{2}(b_3\sigma_x+c_3\tau_{xy})=0.005\times0.8\times(-37.5)=-0.15$ kN, which balances the applied $F_{3x}$.

Answer: $u_3=-2.143\times10^{-7}$ m, $v_3=-4.699\times10^{-8}$ m; $\sigma_x=0.748$, $\sigma_y=-15.70$, $\tau_{xy}=-37.50$ kN/m$^2$ (constant over the element).