---
title: "CST 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-k8"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/cst-stiffness-load-displacement-k8
site: BCT Engineering
---

# CST plate: stiffness, load vector, displacements

Computational Techniques in Civil Engineering · Chapter 4: Finite element method

- Asked in: 2078 Kartik, 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-k8
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4

## Question

Using the plane stress condition, determine the stiffness matrix, nodal load vector and nodal displacement of the given CST element. Take thickness of plate $= 10$ mm, unit weight $= 78.5$ kN/m$^3$, $E = 2\times10^5$ MPa and $G = 105\times10^3$ MPa.

[Figure: right triangular plate with node 1 at the bottom-left (pinned), node 2 at the bottom-right 800 mm from node 1 (roller), and node 3 directly above node 1 at 600 mm; the vertical edge 1-3 carries a distributed load of 5 kN/m$^2$]

## Short answer

Using a single CST element under plane stress, the stiffness matrix is 6×6, the nodal load vector from self-weight and edge pressure is {15, -62.8, 0, -62.8, 15, -62.8} N. With u1=v1=v2=0, the computed nodal displacements are u2 = -2.99×10^-6 mm, u3 = 2.14×10^-5 mm, v3 = -4.71×10^-5 mm.

## Answer

**Assumptions:** plane stress, one CST element. Nodes: 1 $(0,0)$ pinned, 2 $(800,0)$ roller, 3 $(0,600)$. Given $E=2\times10^5$ MPa and $G=105\times10^3$ MPa, the Poisson's ratio follows from $G=\dfrac{E}{2(1+\nu)}$:
$$
\nu=\frac{E}{2G}-1=\frac{2\times10^5}{2\times105\times10^3}-1=-0.0476
$$
Units: N, mm.

#### Step 1: Geometry
- Node 1: $(0,\ 0)$ mm
- Node 2: $(800,\ 0)$ mm
- Node 3: $(0,\ 600)$ mm

Area: $A=\tfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=240000\ \text{mm}^2$

$b_1=y_2-y_3=-600,\ b_2=y_3-y_1=600,\ b_3=y_1-y_2=0$ mm

$c_1=x_3-x_2=-800,\ c_2=x_1-x_3=0,\ c_3=x_2-x_1=800$ mm

#### Step 2: [B] and [D] matrices (plane stress)
$$
[B]=\frac{1}{2A}\begin{bmatrix}-600 & 0 & 600 & 0 & 0 & 0 \\ 0 & -800 & 0 & 0 & 0 & 800 \\ -800 & -600 & 0 & 600 & 800 & 0\end{bmatrix},\quad 2A=480000\ \text{mm}^2
$$
$$
[D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix}=10^{5}\begin{bmatrix}2.005 & -0.095 & 0 \\ -0.095 & 2.005 & 0 \\ 0 & 0 & 1.050\end{bmatrix}\ \text{N/mm}^2
$$

#### Step 3: Element stiffness matrix
$$
[k]=t\,A\,[B]^T[D][B]= 10^{6}\begin{bmatrix}1.452 & 0.477 & -0.752 & -0.525 & -0.700 & 0.048 \\ 0.477 & 1.730 & 0.048 & -0.394 & -0.525 & -1.336 \\ -0.752 & 0.048 & 0.752 & 0 & 0 & -0.048 \\ -0.525 & -0.394 & 0 & 0.394 & 0.525 & 0 \\ -0.700 & -0.525 & 0 & 0.525 & 0.700 & 0 \\ 0.048 & -1.336 & -0.048 & 0 & 0 & 1.336\end{bmatrix}\ \text{N/mm}
$$
(DOF order: $u_1,v_1,u_2,v_2,u_3,v_3$.)

#### Step 4: Load vector
Self-weight $W=\gamma tA=78.5\times10^{-6}\times10\times240000=188.4$ N, i.e. $62.8$ N downward at each node. The 5 kN/m$^2$ = 0.005 N/mm$^2$ on edge 1-3 (600 mm long, acting horizontally toward $+x$, assumed) gives $0.005\times10\times600=30$ N, 15 N to each end node. The full nodal load vector is
$$
\{F\}=\begin{Bmatrix}F_{u1}\\F_{v1}\\F_{u2}\\F_{v2}\\F_{u3}\\F_{v3}\end{Bmatrix}=\begin{Bmatrix}15 \\ -62.8 \\ 0 \\ -62.8 \\ 15 \\ -62.8\end{Bmatrix}\ \text{N}
$$
The loads at the restrained DOFs ($u_1$, $v_1$, $v_2$) go straight into the supports, so the loads on the free DOFs are $F_{u2}=0$, $F_{u3}=15$ N and $F_{v3}=-62.8$ N.

#### Boundary conditions and solution
Node 1 is pinned ($u_1=v_1=0$) and node 2 is a roller ($v_2=0$, $u_2$ free). Unknowns: $u_2,u_3,v_3$ (rows/columns 3, 5, 6).
$$
10^{6}\begin{bmatrix}0.752 & 0 & -0.048 \\ 0 & 0.700 & 0 \\ -0.048 & 0 & 1.336\end{bmatrix}\begin{Bmatrix}u_2 \\ u_3 \\ v_3\end{Bmatrix}=\begin{Bmatrix}0 \\ 15 \\ -62.8\end{Bmatrix}
$$
$$
\begin{aligned} u_2&=-2.99\times10^{-6}\ \text{mm} \\ u_3&=2.143\times10^{-5}\ \text{mm} \\ v_3&=-4.71\times10^{-5}\ \text{mm} \end{aligned}
$$

#### Strains and stresses (constant over the element, so equal at the centroid)
Centroid: $(266.67,\ 200.00)$ mm.
$$
\{\varepsilon\}=[B]\{d\}=\begin{Bmatrix}-3.738\times10^{-9} \\ -7.85\times10^{-8} \\ 3.571\times10^{-8}\end{Bmatrix}
$$
$$
\{\sigma\}=[D]\{\varepsilon\}=\begin{Bmatrix}0 \\ -0.0157 \\ 0.00375\end{Bmatrix}\ \text{MPa}
$$

**Answer:** $[k]$ and $\{F\}$ as above; $u_2=-2.99\times10^{-6}$ mm, $u_3=2.14\times10^{-5}$ mm, $v_3=-4.71\times10^{-5}$ mm, with $u_1=v_1=v_2=0$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is nu found from E and G?
   A. nu = E/(2G) - 1
   B. nu = G/E
   C. nu = 2G/E - 1
   D. nu = E/G
   Answer: A. From G = E/(2(1+nu)); here nu = -0.0476. (Key point: Poisson's ratio from G)

2. Node 1 pinned and node 2 roller. Which are unknown?
   A. u2, u3, v3
   B. u1, v1, v2
   C. u2, v2, u3
   D. Only v3
   Answer: A. u1 = v1 = 0 and v2 = 0; u2, u3, v3 are free. (Key point: Boundary conditions)

3. The 5 kN/m^2 on the 600 mm edge gives a total of
   A. 30 N, 15 N to each end node
   B. 3000 N
   C. 30 N all to node 3
   D. 15 N total
   Answer: A. 0.005 x 10 x 600 = 30 N, half to each end; the 15 N at pinned node 1 goes into the support. (Key point: Edge load distribution)

4. Self-weight is W = gamma t A with A = 240000 mm^2. The nodal share is
   A. 62.8 N downward each
   B. 188.4 N each
   C. 31.4 N each
   D. 0
   Answer: A. W = 188.4 N, divided equally among three nodes = 62.8 N. (Key point: Self-weight per node)

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