--- title: "CST: nodal deformations and stresses" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:cst-plane-stress-m3" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/cst-plane-stress-m3 site: BCT Engineering --- # CST: nodal deformations and stresses Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2073 Magh, 10 marks - Video: [FEA nodal force calculations](https://www.youtube.com/watch?v=o2xzEIlUkuU) — Er Ash mam - 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-plane-stress-m3 - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question A steel plate of 10 mm thickness is loaded as shown in the figure below. For the plane stress problem, obtain the nodal deformations and the stresses in the CST element. Take $E = 2\times10^5$ MPa, $G = 105\times10^3$ MPa and unit weight of steel is 78.5 kN/m$^3$. [Figure: right triangular plate with node 1 at the origin, node 2 at 80 cm to the right of node 1 and node 3 at 60 cm above node 1; a 50 kN/m$^2$ distributed load acts on the left edge 1-3, directed in the negative x direction] ## Short answer Using one CST in plane stress with the given G, node 3 displaces u3=-2.143e-7 m and v3=-4.699e-8 m. The constant element stresses are sigma_x=0.748, sigma_y=-15.70, and tau_xy=-37.50 kN/m^2. ## Answer 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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With E = 2 x 10^8 and G = 105 x 10^6 kN/m^2 used as given, what is nu? A. 0.3 B. E/(2G) - 1 = -0.0476 C. 0.5 D. 0 Answer: B. nu = E/(2G) - 1 = 2/2.1 - 1 = -0.0476 (the data are followed as stated). (Key point: Poisson's ratio from G) 2. Which nodal displacements are unknown? A. All six B. u3 and v3 only, since nodes 1 and 2 on the base are fixed C. u2 and v2 only D. u1 and v3 only Answer: B. The base nodes are fixed, so node 3 is free and gives a 2 x 2 reduced system. (Key point: Boundary conditions) 3. How is the 0.30 kN edge load and the 0.1884 kN self-weight applied? A. Edge load 0.15 kN in -x at node 3 (shared with node 1), self-weight 1/3 to each node giving -0.0628 kN at node 3 B. All 0.30 kN at node 3 C. Self-weight all at node 2 D. The edge load acts in y Answer: A. The edge load is shared equally by its end nodes 1 and 3, and self-weight is split in three. (Key point: Nodal loads) 4. How is tau_xy = -37.5 kN/m^2 verified? A. 0.3 kN carried by shear on the 0.8 m base: 0.3/(0.01 x 0.8) = 37.5 B. Using P/A with A = 0.24 m^2 C. Using self-weight/area D. It cannot be checked Answer: A. The horizontal edge load is resisted by shear over the fixed base of area 0.01 x 0.8 m^2. 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