--- title: "CST steel plate: stiffness, load vector, displacements" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:cst-stiffness-load-displacement-c8" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/cst-stiffness-load-displacement-c8 site: BCT Engineering --- # CST steel plate: stiffness, load vector, displacements Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2078 Chaitra, 4+2+4 marks - Video: [Two Dimensional (2D) CST element problem | Stiffness Matrix | Finite element Analysis | FEA in Tamil](https://www.youtube.com/watch?v=RSp8bE7MNWs) — Focus Academy Lectures - 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-stiffness-load-displacement-c8 - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question A steel plate of uniform thickness 10 mm is being loaded as shown in the figure below. Considering the plane stress condition, determine the stiffness matrix, load vector and nodal displacements of the given CST element. Take modulus of elasticity $= 200\times10^3$ MPa, Poisson's ratio $= 0.3$ and unit weight of steel $= 78.5$ kN/m$^3$. [Figure: right triangular plate, 500 mm wide at the top (fixed along the top edge) and 500 mm deep on the left edge, which carries a distributed load of 5 kN/m$^2$] ## Short answer For the single plane-stress CST element, form [B] and [D], then [k]=t A [B]^T[D][B] as given in N/mm. The load vector has F_x1=12.5 N and F_y1=-32.71 N at node 1, with nodes 2 and 3 fixed. Solving gives u_1=3.25e-5 mm and v_1=-2.98e-5 mm. ## Answer **Assumptions:** plane stress, one CST element. Nodes: 1 = bottom of left edge $(0,-500)$, 2 = top right $(500,0)$, 3 = top left $(0,0)$ (counter-clockwise). $E=200\times10^3$ MPa, $\nu=0.3$, $t=10$ mm, unit weight $78.5\times10^{-6}$ N/mm$^3$. Units: N, mm. #### Step 1: Geometry - Node 1 (bottom of left edge): $(0,\ -500)$ mm - Node 2 (top right): $(500,\ 0)$ mm - Node 3 (top left): $(0,\ 0)$ mm Area: $A=\tfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=125000\ \text{mm}^2$ $b_1=y_2-y_3=0,\ b_2=y_3-y_1=500,\ b_3=y_1-y_2=-500$ mm $c_1=x_3-x_2=-500,\ c_2=x_1-x_3=0,\ c_3=x_2-x_1=500$ mm #### Step 2: [B] and [D] matrices (plane stress) $$ [B]=\frac{1}{2A}\begin{bmatrix}0 & 0 & 500 & 0 & -500 & 0 \\ 0 & -500 & 0 & 0 & 0 & 500 \\ -500 & 0 & 0 & 500 & 500 & -500\end{bmatrix},\quad 2A=250000\ \text{mm}^2 $$ $$ [D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix}=10^{5}\begin{bmatrix}2.198 & 0.659 & 0 \\ 0.659 & 2.198 & 0 \\ 0 & 0 & 0.769\end{bmatrix}\ \text{N/mm}^2 $$ #### Step 3: Element stiffness matrix $$ [k]=t\,A\,[B]^T[D][B]= 10^{6}\begin{bmatrix}0.385 & 0 & 0 & -0.385 & -0.385 & 0.385 \\ 0 & 1.099 & -0.330 & 0 & 0.330 & -1.099 \\ 0 & -0.330 & 1.099 & 0 & -1.099 & 0.330 \\ -0.385 & 0 & 0 & 0.385 & 0.385 & -0.385 \\ -0.385 & 0.330 & -1.099 & 0.385 & 1.484 & -0.714 \\ 0.385 & -1.099 & 0.330 & -0.385 & -0.714 & 1.484\end{bmatrix}\ \text{N/mm} $$ (DOF order: $u_1,v_1,u_2,v_2,u_3,v_3$.) #### Step 4: Load vector Loads: (i) self-weight $W=\gamma tA=78.5\times10^{-6}\times10\times125000=98.125$ N, shared equally: $W/3=32.708$ N downward at each node. (ii) The 5 kN/m$^2$ = 0.005 N/mm$^2$ traction on the left edge 1-3 (length 500 mm, acting horizontally toward $+x$, assumed): total $0.005\times10\times500=25$ N, half (12.5 N) to each end node. $$ \{F\}=\begin{Bmatrix}12.5 \\ -32.71 \\ 0 \\ -32.71 \\ 12.5 \\ -32.71\end{Bmatrix}\ \text{N} $$ #### Boundary conditions and solution The top edge is fixed, so nodes 2 and 3 have zero displacement. The only free node is node 1 ($u_1,v_1$); use rows and columns 1 and 2 of $[k]$. $$ 10^{6}\begin{bmatrix}0.385 & 0 \\ 0 & 1.099\end{bmatrix}\begin{Bmatrix}u_1 \\ v_1\end{Bmatrix}=\begin{Bmatrix}12.5 \\ -32.71\end{Bmatrix} $$ $$ \begin{aligned} u_1&=3.25\times10^{-5}\ \text{mm} \\ v_1&=-2.976\times10^{-5}\ \text{mm} \end{aligned} $$ #### Strains and stresses (constant over the element, so equal at the centroid) Centroid: $(166.67,\ -166.67)$ mm. $$ \{\varepsilon\}=[B]\{d\}=\begin{Bmatrix}0 \\ 5.953\times10^{-8} \\ -6.5\times10^{-8}\end{Bmatrix} $$ $$ \{\sigma\}=[D]\{\varepsilon\}=\begin{Bmatrix}0.003925 \\ 0.01308 \\ -0.005\end{Bmatrix}\ \text{MPa} $$ **Answer:** $[k]$ as above (units N/mm); nodal loads $\{F\}$ as above (only node 1 matters: $F_{x1}=12.5$ N, $F_{y1}=-32.71$ N); $u_1=3.25\times10^{-5}$ mm, $v_1=-2.98\times10^{-5}$ mm. The plate is very stiff, so the displacements are extremely small. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the self-weight distributed to the nodes of the CST? A. W = gamma t A shared equally, W/3 at each node B. All at node 1 C. In proportion to the coordinates D. It is ignored Answer: A. W = 78.5e-6 x 10 x 125000 = 98.125 N, giving 32.708 N per node. (Key point: Self-weight load) 2. Why are only rows and columns 1 and 2 of [k] used? A. The top edge (nodes 2 and 3) is fixed, so only node 1 can move B. The load acts only at node 1 C. [k] is zero elsewhere D. Node 1 is fixed Answer: A. Fixed nodes have zero displacements, leaving u1 and v1. (Key point: Free node) 3. How is the 5 kN/m^2 on the 500 mm edge converted to nodal loads? A. Total 0.005 x 10 x 500 = 25 N, half (12.5 N) to each end node B. 25 N to node 1 only C. 5000 N at each node D. 12.5 N at all three nodes Answer: A. Traction x thickness x length gives 25 N, shared equally between the edge's end nodes. (Key point: Edge load to nodes) 4. Why are the computed displacements so small (about 10^-5 mm)? A. Loads are tiny relative to the very stiff steel plate B. Because of an error in [D] C. Plane stress reduces displacement D. The mesh is too fine Answer: A. Loads of tens of newtons on a steel plate with E = 200 GPa and a large area produce negligible movement. 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