--- title: "CST: nodal displacements and stresses (psi)" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:cst-shape-function-nodal-displacements-b0" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/cst-shape-function-nodal-displacements-b0 site: BCT Engineering --- # CST: nodal displacements and stresses (psi) Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2070 Bhadra, 12 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-shape-function-nodal-displacements-b0 - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question Considering the plane stress condition, find out the nodal displacements and stresses of the CST element as shown in Fig. 2. $E = 30\times10^6$ psi, $t = 0.3$ in, $\gamma = 460$ lb/in$^3$, $\nu = 0.3$, $T_3 = 360$ psi with usual notations. [Figure: triangular element with nodes 1 and 2 at the base (pin supports at both) and node 3 at the top; a distributed traction $T_3$ acts normal to the inclined edge 2-3] ## Short answer Use one CST in plane stress with nodes 1 and 2 pinned; convert traction and self weight into nodal loads at node 3. Displacements are u3=-1.404e-4 in and v3=-2.330e-4 in. Stresses are sigma_x=-384 psi, sigma_y=-1280 psi, tau_xy=-270 psi. ## Answer 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Nodes 1 and 2 are pinned and node 3 is free. How many unknown nodal displacements are solved? A. 2 (u_3 and v_3) B. 6 C. 4 D. 3 Answer: A. Only node 3 can move, so the unknowns are u_3 and v_3 and the equation uses the 2 x 2 block [k]_33. (Key point: Free dofs and supports) 2. How is the edge traction T = 360 psi on the 10 in edge 2-3 converted to nodal loads? A. Total T t L = 1080 lb is lumped entirely at node 3 B. The full 360 lb is applied at each node C. Total 1080 lb is split equally, 540 lb to each end node, and resolved with the edge normal D. T times area of the triangle, 360 x 24 lb Answer: C. Force on the edge is T t L = 1080 lb, shared half to each end node, then resolved along the normal (0.6, 0.8). (Key point: Traction load) 3. How is the self weight included at node 3? A. The full weight 3312 lb acts at node 3 B. W = gamma t A = 3312 lb and one third (1104 lb) is placed downward at node 3 C. Half of the weight acts at node 3 D. Self weight is ignored for a plane stress element Answer: B. A CST body force is shared equally among its three nodes, so node 3 gets W/3 = 1104 lb in -y. (Key point: Self weight) 4. Once u_3 and v_3 are known, how are the element stresses found? A. sigma = [B]{u} directly B. sigma = [D]{u}/A C. sigma = [k]{u}/t D. sigma = [D][B]{u} Answer: D. Strain is [B]{u}, and stress is [D] times strain, giving sigma_x, sigma_y and tau_xy as constants in the CST. 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