CST: nodal displacements and stresses (psi)

The element is analysed as one CST in plane stress. Units: lb and in.

Assumptions (the figure gives no dimensions)

• Node 1 (0, 0), node 2 (8, 0), node 3 (0, 6) in, so the inclined edge 2-3 is 10 in long with unit normal $(0.6,\ 0.8)$. Nodes 1 and 2 are pinned ($u=v=0$); node 3 is free.
• $T_3=360$ psi is a pressure pushing on edge 2-3 towards the element. Self weight acts in $-y$ with $\gamma=460$ lb/in$^3$ as given.
• $t=0.3$ in, $E=30\times10^6$ psi, $\nu=0.3$.

Step 1: Geometry and [B]

$$A=\tfrac12(8)(6)=24\ \text{in}^2,\qquad [B]=\frac{1}{48}\begin{bmatrix}-6&0&6&0&0&0\\0&-8&0&0&0&8\\-8&-6&0&6&8&0\end{bmatrix}$$

Step 2: [D] and [k]

$$[D]=\begin{bmatrix} 3.2967\times10^{7} & 9890110 & 0 \\ 9890110 & 3.2967\times10^{7} & 0 \\ 0 & 0 & 1.1538\times10^{7} \end{bmatrix}\ \text{psi}$$

Stiffness of the free node 3 ($[k]=tA[B]^T[D][B]$):

$$[k]_{33}=\begin{bmatrix} 2307692 & 0 \\ 0 & 6593407 \end{bmatrix}\ \text{lb/in}$$

Step 3: Nodal loads at node 3

• Traction: total force $=T\,t\,L=360\times0.3\times10=1080$ lb, half to each end node of edge 2-3, i.e. 540 lb at node 3. Its direction is opposite to the outward normal, so $F_x=-540(0.6)=-324$ lb and $F_y=-540(0.8)=-432$ lb.
• Self weight: $W=\gamma tA=460\times0.3\times24=3312$ lb; one third at node 3 is $1104$ lb downward.

$$\{F_3\}=\begin{bmatrix}-324\\-432-1104\end{bmatrix}=\begin{bmatrix}-324\\-1536\end{bmatrix}\text{lb}$$

Step 4: Nodal displacements

$$[k]_{33}\begin{bmatrix}u_3\\v_3\end{bmatrix}=\begin{bmatrix}-324\\-1536\end{bmatrix}\ \Rightarrow\ u_3=-1.404\times10^{-4}\ \text{in},\ v_3=-2.330\times10^{-4}\ \text{in}$$

$u_1=v_1=u_2=v_2=0$.

Step 5: Stresses

$$\{\sigma\}=[D][B]\{u\}=\begin{bmatrix}-384\\-1280\\-270\end{bmatrix}\ \text{psi}$$

Answer: $u_3=-1.404\times10^{-4}$ in, $v_3=-2.330\times10^{-4}$ in; $\sigma_x=-384$ psi, $\sigma_y=-1280$ psi, $\tau_{xy}=-270$ psi. If the actual figure has other dimensions, only the numbers in Steps 1-5 change; the method is the same.