--- title: "Thin plate: nodal displacements and stresses" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:cst-thin-plate-psi" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/cst-thin-plate-psi site: BCT Engineering --- # Thin plate: nodal displacements and stresses Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2071 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/cst-thin-plate-psi - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question A thin plate is subjected to the loads as shown in the figure below. The plate thickness is 0.3 in and the other dimensions are shown in the figure. Given that the Poisson's ratio $= 0.3$ and the modulus of elasticity $E = 30\times10^6$ psi, determine the nodal load displacements and the elemental stresses. [Figure: triangular plate with node 1 at the bottom-left (pinned), node 2 directly above node 1 at 10 in (supported), node 3 at 20 in to the right of node 1 and 10 in above it; at node 3 a 2000 lb horizontal load (to the right) and a 2000 lb downward load] ## Short answer Model as one CST in plane stress with node 1 pinned and node 2 restrained horizontally. Free displacements are v2=-3.556e-4 in, u3=1.022e-3 in, v3=-2.667e-3 in. Element stresses are sigma_x=1333.3 psi, sigma_y=-666.7 psi, tau_xy=-1333.3 psi. ## Answer The plate is a single CST in plane stress. Units: lb and in. #### Assumptions - Node coordinates (in): 1 (0, 0), 2 (0, 10), 3 (20, 10). Nodes are taken in anticlockwise order **1-3-2** ($i=1,\ j=3,\ m=2$) so that the area is positive. - Node 1 is pinned ($u_1=v_1=0$); node 2 is on a roller that stops horizontal movement ($u_2=0$). Free dofs: $v_2,\ u_3,\ v_3$. - Loads at node 3: $F_{3x}=+2000$ lb, $F_{3y}=-2000$ lb. - $E=30\times10^6$ psi, $\nu=0.3$, $t=0.3$ in. #### Step 1: Geometry $$ A=\tfrac12(20)(10)=100\ \text{in}^2 $$ $b_i=y_j-y_m=0,\ b_j=y_m-y_i=10,\ b_m=y_i-y_j=-10$; $c_i=x_m-x_j=-20,\ c_j=x_i-x_m=0,\ c_m=x_j-x_i=20$. $$ [B]=\frac{1}{200}\begin{bmatrix}0&0&10&0&-10&0\\0&-20&0&0&0&20\\-20&0&0&10&20&-10\end{bmatrix} $$ (columns in the order $u_1,v_1,u_3,v_3,u_2,v_2$). #### Step 2: Elasticity matrix $$ [D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix}=\begin{bmatrix} 3.2967\times10^{7} & 9890110 & 0 \\ 9890110 & 3.2967\times10^{7} & 0 \\ 0 & 0 & 1.1538\times10^{7} \end{bmatrix}\ \text{psi} $$ #### Step 3: Stiffness and reduced equations $[k]=tA[B]^T[D][B]$ is assembled in global dof order $(u_1,v_1,u_2,v_2,u_3,v_3)$. Removing the rows and columns of $u_1,v_1,u_2$: $$ \begin{bmatrix} 1.0755\times10^{7} & 1483516 & -865385 \\ 1483516 & 2472527 & 0 \\ -865385 & 0 & 865385 \end{bmatrix}\begin{bmatrix}v_2\\u_3\\v_3\end{bmatrix}=\begin{bmatrix}0\\2000\\-2000\end{bmatrix} $$ #### Step 4: Nodal displacements $$ v_2=-3.556\times10^{-4}\ \text{in},\quad u_3=1.022\times10^{-3}\ \text{in},\quad v_3=-2.667\times10^{-3}\ \text{in} $$ Reactions (check): $R_{1x}=4000$, $R_{1y}=2000$, $R_{2x}=-6000$ lb. Then $\Sigma F_x = 4000-6000+2000=0$ and $\Sigma F_y = 2000-2000=0$. #### Step 5: Element stresses, $\{\sigma\}=[D][B]\{u\}$ $$ \sigma_x=1333.3\ \text{psi},\quad \sigma_y=-666.7\ \text{psi},\quad \tau_{xy}=-1333.3\ \text{psi} $$ Principal stresses: $\sigma_1=2000.0$ psi, $\sigma_2=-1333.3$ psi, at $\theta_p=-26.57^\circ$ from the $x$-axis. **Answer:** $v_2=-3.556\times10^{-4}$ in, $u_3=1.022\times10^{-3}$ in, $v_3=-2.667\times10^{-3}$ in; $\sigma_x=1333.3$, $\sigma_y=-666.7$, $\tau_{xy}=-1333.3$ psi. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are the nodes taken in the order 1-3-2 for the CST? A. So that the node with the load comes first B. So that the stiffness matrix becomes symmetric C. So that the nodes are anticlockwise and the area is positive D. So that the roller node is last Answer: C. The area formula needs anticlockwise ordering to give a positive area, so i = 1, j = 3, m = 2 is used. (Key point: Node ordering) 2. Node 1 is pinned and node 2 is a roller stopping horizontal movement. Which dofs are free? A. u_2, u_3, v_3 B. v_1, v_2, u_3 C. v_2, u_3, v_3 D. u_1, u_3, v_3 Answer: C. u_1 = v_1 = 0 and u_2 = 0 are restrained, leaving v_2, u_3 and v_3. (Key point: Free dofs) 3. Which expression gives the stiffness matrix and the stresses of the CST in plane stress? A. [k] = tA[B]^T[D][B] and stress = [D][B]{u} B. [k] = [B]^T[D][B]/(tA) and stress = [B]{u} C. [k] = t[B][D][B]^T and stress = [D]{u} D. [k] = AE/L [B] and stress = E[B]{u} Answer: A. The CST has constant [B], so the stiffness is tA[B]^T[D][B] and the stress is [D][B]{u}. (Key point: CST matrices) 4. How are the computed displacements checked? A. By requiring all stresses equal B. By finding reactions (R1x = 4000, R1y = 2000, R2x = -6000 lb) and confirming sum of forces is zero in x and y C. By checking v_2 equals v_3 D. By confirming that D is symmetric Answer: B. The reactions together with the 2000 lb loads give zero resultant in both x and y, confirming the solution. 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