--- title: "CST: stiffness, nodal displacement, stresses" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:cst-plane-stress-b5" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/cst-plane-stress-b5 site: BCT Engineering --- # CST: stiffness, nodal displacement, stresses Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2075 Bhadra, 10 marks - Video: [FEA part1 Displacement stress reaction numerical](https://www.youtube.com/watch?v=MbHeAEceKBE) — 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-b5 - 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 for this CST element, determine (a) element stiffness matrix, (b) nodal displacements, and (c) strains and stresses at the centroid of the element. Take $E = 210\times10^3$ MPa and $G = 105\times10^3$ MPa. The unit weight of steel is 78.5 kN/m$^3$. [Figure: right triangular plate; node 1 at the top-left, node 2 at the bottom of the left edge (300 mm below node 1, supports along the left edge), node 3 at the top-right 400 mm from node 1; a distributed load of 25 kN/m acts downward along the top edge 1-3] ## Short answer Plane stress CST with E=210 GPa gives nu=0. Using [k]=tA[B]^T[D][B] and the left edge fixed, free DOFs are u3=0 mm and v3=-0.01274 mm. Centroid strains are epsilon_x=0, epsilon_y=0, gamma_xy=-3.185e-5; stresses are sigma_x=0, sigma_y=0, tau_xy=-3.344 MPa. ## Answer **Data:** plane stress, one CST element, $t=10$ mm, $E=210\times10^3$ MPa. From $G=E/[2(1+\nu)]=105\times10^3$ MPa, $\nu=E/(2G)-1=0$. Nodes: 1 = top left $(0,0)$, 2 = bottom of left edge $(0,-300)$, 3 = top right $(400,0)$. Units: N, mm. #### Step 1: Geometry - Node 1 (top left): $(0,\ 0)$ mm - Node 2 (bottom left): $(0,\ -300)$ mm - Node 3 (top right): $(400,\ 0)$ mm Area: $A=\tfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=60000\ \text{mm}^2$ $b_1=y_2-y_3=-300,\ b_2=y_3-y_1=0,\ b_3=y_1-y_2=300$ mm $c_1=x_3-x_2=400,\ c_2=x_1-x_3=-400,\ c_3=x_2-x_1=0$ mm #### Step 2: [B] and [D] matrices (plane stress) $$ [B]=\frac{1}{2A}\begin{bmatrix}-300 & 0 & 0 & 0 & 300 & 0 \\ 0 & 400 & 0 & -400 & 0 & 0 \\ 400 & -300 & -400 & 0 & 0 & 300\end{bmatrix},\quad 2A=120000\ \text{mm}^2 $$ $$ [D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix}=10^{5}\begin{bmatrix}2.100 & 0 & 0 \\ 0 & 2.100 & 0 \\ 0 & 0 & 1.050\end{bmatrix}\ \text{N/mm}^2 $$ #### Step 3: Element stiffness matrix $$ [k]=t\,A\,[B]^T[D][B]= 10^{6}\begin{bmatrix}1.488 & -0.525 & -0.700 & 0 & -0.787 & 0.525 \\ -0.525 & 1.794 & 0.525 & -1.400 & 0 & -0.394 \\ -0.700 & 0.525 & 0.700 & 0 & 0 & -0.525 \\ 0 & -1.400 & 0 & 1.400 & 0 & 0 \\ -0.787 & 0 & 0 & 0 & 0.787 & 0 \\ 0.525 & -0.394 & -0.525 & 0 & 0 & 0.394\end{bmatrix}\ \text{N/mm} $$ (DOF order: $u_1,v_1,u_2,v_2,u_3,v_3$.) #### Step 4: Load vector Top edge load $25$ kN/m $=25$ N/mm over $400$ mm $=10\,000$ N, shared: $5000$ N at node 1 (absorbed by the support) and $5000$ N at node 3. Self-weight $W=78.5\times10^{-6}\times10\times60000=47.1$ N, $15.70$ N downward at each node. Load on the free node 3: $F_{y3}=-(5000+15.70)=-5015.70$ N. $$ \{F\}=\begin{Bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ -5016\end{Bmatrix}\ \text{N} $$ #### Boundary conditions and solution Nodes 1 and 2 lie on the supported left edge ($u=v=0$). Free DOFs: $u_3,v_3$ (rows/columns 5 and 6). $$ 10^{5}\begin{bmatrix}7.875 & 0 \\ 0 & 3.938\end{bmatrix}\begin{Bmatrix}u_3 \\ v_3\end{Bmatrix}=\begin{Bmatrix}0 \\ -5016\end{Bmatrix} $$ $$ \begin{aligned} u_3&=0\ \text{mm} \\ v_3&=-0.01274\ \text{mm} \end{aligned} $$ #### Strains and stresses (constant over the element, so equal at the centroid) Centroid: $(133.33,\ -100.00)$ mm. $$ \{\varepsilon\}=[B]\{d\}=\begin{Bmatrix}0 \\ 0 \\ -3.185\times10^{-5}\end{Bmatrix} $$ $$ \{\sigma\}=[D]\{\varepsilon\}=\begin{Bmatrix}0 \\ 0 \\ -3.344\end{Bmatrix}\ \text{MPa} $$ **Answer:** (a) element stiffness matrix as given in Step 3 (N/mm); (b) $u_3=0$ mm, $v_3=-0.0127$ mm (downward), others zero; (c) at the centroid $\varepsilon_x=0$, $\varepsilon_y=0$, $\gamma_{xy}=-3.18\times10^{-5}$ and $\sigma_x=0.000$, $\sigma_y=0.000$, $\tau_{xy}=-3.344$ N/mm$^2$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With E = 210 x 10^3 MPa and G = 105 x 10^3 MPa, what is nu and what does it do to [D]? A. nu = 0.3, so [D] has coupling terms B. nu = 0, since G = E/[2(1+nu)] gives E/(2G) - 1 = 0, so [D] is diagonal C. nu = 0.5, so the material is incompressible D. nu = 1, so [D] is singular Answer: B. E/(2G) = 1 so nu = 0, and [D] becomes diag(E, E, E/2). (Key point: Poisson's ratio from G) 2. For nodes (0,0), (0,-300), (400,0), what are A and b1 = y2 - y3? A. A = 120000 mm^2, b1 = 300 B. A = 60000 mm^2, b1 = -300 C. A = 60000 mm^2, b1 = 400 D. A = 30000 mm^2, b1 = -300 Answer: B. A = 1/2 x 400 x 300 = 60000 mm^2 and b1 = -300 - 0 = -300. (Key point: Area and B coefficients) 3. How is the 25 kN/m edge load on edge 1-3 applied to the nodes? A. The full 10 000 N goes to node 3 B. 5000 N each to nodes 1 and 3; node 1 is supported, so only node 3 carries load (plus self-weight share) C. 10 000 N to node 2 D. It is ignored because the edge is horizontal Answer: B. The 10 000 N resultant is shared equally; node 3 gets 5000 + 15.7 = 5015.7 N downward. (Key point: Load vector) 4. Why are the strains and stresses at the centroid equal to those found anywhere in the element? A. The centroid is the only point where they can be computed B. [B] contains only nodal coordinates, so strain and stress are constant over a CST C. The load acts at the centroid D. The plate has zero thickness variation Answer: B. [B] is constant for a CST, so e = B d and s = D e are uniform throughout. 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