--- title: "CST equilateral plate: stresses at centroid" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:cst-equilateral-10kn-edge" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/cst-equilateral-10kn-edge site: BCT Engineering --- # CST equilateral plate: stresses at centroid Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2074 Bhadra, 10 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-equilateral-10kn-edge - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question A steel plate of thickness 10 mm is being loaded as shown in the figure below. Considering the plane stress condition, determine the stresses and strains at the centroid of the CST element. Take $E = 210\times10^3$ MPa, $\nu = 0.30$ and unit weight of steel is 78.50 kN/m$^3$, length of each side $= 100$ mm. [Figure: equilateral triangle with nodes 1 (0, 0) and 2 (100 mm, 0) on the supported base (50 mm each side of the y-axis) and node 3 at the apex; a distributed load of 10 kN/m acts on the inclined edge 2-3] ## Short answer Plane stress CST with sides 100 mm, nu=0.3. With base nodes supported and load plus self-weight at node 3, free DOFs are u3=0 mm and v3=-3.761e-4 mm. Centroid strains are epsilon_x=0, epsilon_y=-4.343e-6, gamma_xy=0; stresses are sigma_x=-0.301 MPa, sigma_y=-1.002 MPa, tau_xy=0. ## Answer **Data:** plane stress, one CST element, side $=100$ mm, $t=10$ mm, $E=210\times10^3$ MPa, $\nu=0.3$, unit weight $78.5\times10^{-6}$ N/mm$^3$. Nodes: 1 $(0,0)$, 2 $(100,0)$, 3 $(50,\ 86.603)$. Units: N, mm. #### Step 1: Geometry - Node 1: $(0,\ 0)$ mm - Node 2: $(100,\ 0)$ mm - Node 3: $(50,\ 86.6025)$ mm Area: $A=\tfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=4330.13\ \text{mm}^2$ $b_1=y_2-y_3=-86.6025,\ b_2=y_3-y_1=86.6025,\ b_3=y_1-y_2=0$ mm $c_1=x_3-x_2=-50,\ c_2=x_1-x_3=-50,\ c_3=x_2-x_1=100$ mm #### Step 2: [B] and [D] matrices (plane stress) $$ [B]=\frac{1}{2A}\begin{bmatrix}-86.60 & 0 & 86.60 & 0 & 0 & 0 \\ 0 & -50 & 0 & -50 & 0 & 100 \\ -50 & -86.60 & -50 & 86.60 & 100 & 0\end{bmatrix},\quad 2A=8660.25\ \text{mm}^2 $$ $$ [D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix}=10^{5}\begin{bmatrix}2.308 & 0.692 & 0 \\ 0.692 & 2.308 & 0 \\ 0 & 0 & 0.808\end{bmatrix}\ \text{N/mm}^2 $$ #### Step 3: Element stiffness matrix $$ [k]=t\,A\,[B]^T[D][B]= 10^{6}\begin{bmatrix}1.116 & 0.375 & -0.883 & -0.029 & -0.233 & -0.346 \\ 0.375 & 0.683 & 0.029 & -0.017 & -0.404 & -0.666 \\ -0.883 & 0.029 & 1.116 & -0.375 & -0.233 & 0.346 \\ -0.029 & -0.017 & -0.375 & 0.683 & 0.404 & -0.666 \\ -0.233 & -0.404 & -0.233 & 0.404 & 0.466 & 0 \\ -0.346 & -0.666 & 0.346 & -0.666 & 0 & 1.332\end{bmatrix}\ \text{N/mm} $$ (DOF order: $u_1,v_1,u_2,v_2,u_3,v_3$.) #### Step 4: Load vector The 10 kN/m on the inclined edge 2-3 (length 100 mm) gives a total of $10\times0.1=1$ kN $=1000$ N. Assumed to act vertically downward, it is shared equally: 500 N at node 2 (absorbed by the support) and 500 N at node 3. Self-weight $W=78.5\times10^{-6}\times10\times4330.1=3.399$ N, $1.133$ N downward at each node. Load on the free node: $F_{y3}=-501.13$ N. $$ \{F\}=\begin{Bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ -501.1\end{Bmatrix}\ \text{N} $$ #### Boundary conditions and solution Nodes 1 and 2 are on the supported base ($u=v=0$). Free DOFs: $u_3,v_3$ (rows/columns 5 and 6). $$ 10^{6}\begin{bmatrix}0.466 & 0 \\ 0 & 1.332\end{bmatrix}\begin{Bmatrix}u_3 \\ v_3\end{Bmatrix}=\begin{Bmatrix}0 \\ -501.1\end{Bmatrix} $$ $$ \begin{aligned} u_3&=0\ \text{mm} \\ v_3&=-3.761\times10^{-4}\ \text{mm} \end{aligned} $$ #### Strains and stresses (constant over the element, so equal at the centroid) Centroid: $(50.00,\ 28.87)$ mm. $$ \{\varepsilon\}=[B]\{d\}=\begin{Bmatrix}0 \\ -4.343\times10^{-6} \\ 0\end{Bmatrix} $$ $$ \{\sigma\}=[D]\{\varepsilon\}=\begin{Bmatrix}-0.3007 \\ -1.002 \\ 0\end{Bmatrix}\ \text{MPa} $$ **Answer:** $u_3=0$ mm, $v_3=-3.76\times10^{-4}$ mm. At the centroid $\varepsilon_x=0$, $\varepsilon_y=-4.34\times10^{-6}$, $\gamma_{xy}=0$; $\sigma_x=-0.301$, $\sigma_y=-1.002$, $\tau_{xy}=0.000$ N/mm$^2$ (compressive). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the equilateral triangle of side 100 mm, what are the apex height and area? A. Height 100 mm, area 5000 mm^2 B. Height 86.603 mm, area 4330.13 mm^2 C. Height 70.7 mm, area 3535 mm^2 D. Height 86.603 mm, area 8660 mm^2 Answer: B. Height = 100 sin 60 deg = 86.603 mm and A = 1/2 x 100 x 86.603 = 4330.13 mm^2. (Key point: Geometry) 2. How is the 10 kN/m on inclined edge 2-3 (100 mm long) converted to nodal loads? A. 1000 N total, split 500 N each to nodes 2 and 3; node 2 is supported so only node 3 matters (plus 1.133 N self-weight) B. 10 kN at node 3 C. 1000 N to node 1 D. It acts only on the support Answer: A. Total = 10 x 0.1 = 1 kN = 1000 N, shared equally; F_y3 = -(500 + 1.133) = -501.13 N. (Key point: Load vector) 3. Which DOFs are solved, and what is the result? A. All six; v3 = 0 B. u3 and v3 only, since nodes 1 and 2 are on the supported base; u3 = 0 and v3 is a small negative value C. u1 and v1 only D. v2 and v3 only Answer: B. Only the free node 3 moves, giving u3 = 0 and v3 = -3.76 x 10^-4 mm. (Key point: Solving for free DOFs) 4. Although eps_x = 0, why is sigma_x = -0.301 MPa nonzero? A. A numerical error B. Because nu = 0.3 couples the strains in [D]: sigma_x = D11 eps_x + D12 eps_y C. Because shear strain is nonzero D. Because of self-weight in x Answer: B. sigma_x = 0.692 x 10^5 x (-4.343 x 10^-6) = -0.30 MPa through the D12 Poisson term. 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