---
title: "Strains and stresses in a CST element"
subject: "Finite Element Method"
chapter: 6
key: "semester_8/finite_element_method:cst-strain-stress-calculation"
source: https://www.bctengineer.com/q/semester_8/finite_element_method/cst-strain-stress-calculation
site: BCT Engineering
---

# Strains and stresses in a CST element

Finite Element Method · Chapter 6: Applications in Solid Mechanics

- Asked in: year not given, 8 marks
- Video: [Solved Example: CST element](https://www.youtube.com/watch?v=S_xNDk2-HcA) — Online Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/finite_element_method/cst-strain-stress-calculation
- Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-6

## Question

A CST element in plane stress has nodes 1 (0, 0), 2 (40, 0) and 3 (0, 30), coordinates in mm. $E = 210$ GPa, $\nu = 0.25$ and thickness $t = 10$ mm. The nodal displacements (mm) are $u_1 = 0$, $v_1 = 0$, $u_2 = 0.02$, $v_2 = 0.005$, $u_3 = 0.004$, $v_3 = -0.01$. Find (a) the area and the $[B]$ matrix, (b) the strains, (c) the stresses and (d) the principal stresses and their orientation.

## Short answer

Area A=600 mm^2. Strains are epsilon_x=5.0e-4, epsilon_y=-3.33e-4, gamma_xy=2.58e-4. Stresses: sigma_x=93.3 MPa, sigma_y=-46.7 MPa, tau_xy=21.7 MPa. Principal stresses are sigma_1=96.6 MPa, sigma_2=-50.0 MPa, oriented at 8.6 degrees from the x-axis.

## Answer

#### (a) Area and $[B]$

$$
2A = x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0 + 40(30) + 0 = 1200 \Rightarrow A = 600\ \text{mm}^2
$$

| $i$ | $b_i = y_j - y_k$ | $c_i = x_k - x_j$ |
|---|---|---|
| 1 | $0 - 30 = -30$ | $0 - 40 = -40$ |
| 2 | $30 - 0 = 30$ | $0 - 0 = 0$ |
| 3 | $0 - 0 = 0$ | $40 - 0 = 40$ |

$$
[B] = \frac{1}{1200}\begin{bmatrix} -30 & 0 & 30 & 0 & 0 & 0 \\ 0 & -40 & 0 & 0 & 0 & 40 \\ -40 & -30 & 0 & 30 & 40 & 0 \end{bmatrix}
$$

#### (b) Strains

$$
\begin{aligned}
\varepsilon_x &= \frac{30(0.02)}{1200} = 5.0\times10^{-4} \\
\varepsilon_y &= \frac{40(-0.01) - 40(0)}{1200} = -3.33\times10^{-4} \\
\gamma_{xy} &= \frac{30(0.005) + 40(0.004)}{1200} = \frac{0.31}{1200} = 2.58\times10^{-4}
\end{aligned}
$$

#### (c) Stresses (plane stress)

$\dfrac{E}{1 - \nu^2} = \dfrac{210\,000}{0.9375} = 224\,000$ MPa and $G = \dfrac{E}{2(1 + \nu)} = 84\,000$ MPa.
$$
\begin{aligned}
\sigma_x &= 224\,000\,(5.0\times10^{-4} + 0.25(-3.33\times10^{-4})) = 93.3\ \text{MPa} \\
\sigma_y &= 224\,000\,(-3.33\times10^{-4} + 0.25(5.0\times10^{-4})) = -46.7\ \text{MPa} \\
\tau_{xy} &= 84\,000\,(2.58\times10^{-4}) = 21.7\ \text{MPa}
\end{aligned}
$$

#### (d) Principal stresses

$$
\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = 23.3 \pm \sqrt{70.0^2 + 21.7^2} = 23.3 \pm 73.3
$$
$$
\sigma_1 = 96.6\ \text{MPa},\quad \sigma_2 = -50.0\ \text{MPa}
$$
$$
\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} = \frac{43.4}{140} = 0.310 \Rightarrow \theta_p = 8.6^\circ
$$

**Answer:** $\varepsilon = (5.0\times10^{-4},\ -3.33\times10^{-4},\ 2.58\times10^{-4})$; $\sigma_x = 93.3$ MPa, $\sigma_y = -46.7$ MPa, $\tau_{xy} = 21.7$ MPa; $\sigma_1 = 96.6$ MPa, $\sigma_2 = -50.0$ MPa at $8.6^\circ$ from the x-axis.

## Check yourself

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

1. For nodes (0,0), (40,0), (0,30), what are the area and b1?
   A. A = 600 mm^2, b1 = -30
   B. A = 1200 mm^2, b1 = -30
   C. A = 600 mm^2, b1 = -40
   D. A = 300 mm^2, b1 = 30
   Answer: A. 2A = 1200 so A = 600 mm^2; b1 = y2 - y3 = 0 - 30 = -30. (Key point: Area)

2. How is gamma_xy computed from [B] and what is its value?
   A. (sum of c_i v_i)/2A = -3.33 x 10^-4
   B. (sum of b_i u_i)/2A = 5 x 10^-4
   C. (sum of b_i v_i - c_i u_i)/2A = 1 x 10^-4
   D. (sum of c_i u_i + b_i v_i)/2A = 2.58 x 10^-4
   Answer: D. gamma_xy = [30(0.005) + 40(0.004)]/1200 = 2.58 x 10^-4. (Key point: Strains)

3. Which constants are used to get stresses for plane stress with E = 210 GPa, nu = 0.25?
   A. E = 210,000 MPa and G = 105,000 MPa
   B. E/(1+nu)(1-2nu) = 420,000 MPa
   C. E/(1 - nu) = 280,000 MPa
   D. E/(1 - nu^2) = 224,000 MPa and G = 84,000 MPa
   Answer: D. 210000/0.9375 = 224,000 MPa and G = 210000/2.5 = 84,000 MPa, giving sigma_x = 93.3, sigma_y = -46.7, tau_xy = 21.7 MPa. (Key point: Stresses)

4. Which formula gives the principal stress orientation?
   A. tan 2 theta = tau_xy/(sigma_x + sigma_y), giving 25°
   B. tan theta = tau_xy/sigma_x, giving 13°
   C. tan 2 theta = (sigma_x - sigma_y)/2 tau_xy, giving 81°
   D. tan 2 theta = 2 tau_xy/(sigma_x - sigma_y), giving 8.6°
   Answer: D. tan 2 theta = 43.4/140 = 0.310, so theta = 8.6°, with sigma_1 = 96.6 MPa and sigma_2 = -50 MPa. (Key point: Principal stresses)

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