1. Strain-Displacement Relation for CST
At any point in the Constant Strain Triangle (CST) element used for plane stress or plane strain problems, there are three strain components: $\varepsilon_x$, $\varepsilon_y$, and $\gamma_{xy}$.
The CST element has 3 nodes with 2 DOF each ($u$, $v$), giving 6 DOF total per element.
CST Element — Triangular Element with 3 Nodes (6 DOF)
x
y
O
1
2
3
4
5
1
2
3
4
Edge 1-2
1
(x₁,y₁)
u₁
v₁
2
(x₂,y₂)
3
(x₃,y₃)
Node with 2 DOF (u, v)
Displacement DOF
Plane Stress / Strain
Fig. 1 — CST Element: 3 Nodes, 6 DOF, with node coordinates and displacement DOFs
1.1 Strain Vector
The three strain components are defined as:
$$\{\varepsilon\} = \begin{Bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{Bmatrix} = \begin{Bmatrix} \dfrac{\partial u}{\partial x} \\[6pt] \dfrac{\partial v}{\partial y} \\[6pt] \dfrac{\partial v}{\partial x} + \dfrac{\partial u}{\partial y} \end{Bmatrix} \quad \cdots\text{(i)}$$
1.2 Displacement Interpolation
The displacement at any point is interpolated from nodal displacements using shape functions $N_1, N_2, N_3$:
$$u = \begin{bmatrix} N_1 & N_2 & N_3 & 0 & 0 & 0 \end{bmatrix} \begin{Bmatrix} u_1 \\ u_2 \\ u_3 \\ v_1 \\ v_2 \\ v_3 \end{Bmatrix}$$
$$v = \begin{bmatrix} 0 & 0 & 0 & N_1 & N_2 & N_3 \end{bmatrix} \begin{Bmatrix} u_1 \\ u_2 \\ u_3 \\ v_1 \\ v_2 \\ v_3 \end{Bmatrix} \quad \cdots\text{(ii)}$$
1.3 Shape Functions
Shape functions vary linearly over the element and satisfy $N_i = 1$ at node $i$ and $N_i = 0$ at all other nodes:
Shape Functions ($A$ = area of triangle)
$$N_1 = \frac{a_1 + b_1 x + c_1 y}{2A}, \qquad N_2 = \frac{a_2 + b_2 x + c_2 y}{2A}, \qquad N_3 = \frac{a_3 + b_3 x + c_3 y}{2A}$$
Coefficients (notation: $y_{ij} = y_i - y_j$, $x_{ij} = x_i - x_j$)
$a_i$ $b_i$ $c_i$ i = 1 $x_2 y_3 - x_3 y_2$ $y_2 - y_3 = y_{23}$ $x_3 - x_2 = x_{32}$ i = 2 $x_3 y_1 - x_1 y_3$ $y_3 - y_1 = y_{31}$ $x_1 - x_3 = x_{13}$ i = 3 $x_1 y_2 - x_2 y_1$ $y_1 - y_2 = y_{12}$ $x_2 - x_1 = x_{21}$
Shape Functions for CST Element — N₁, N₂, N₃
N₁
= (a₁+b₁x+c₁y)/2A
1
N₁=1
N₂
= (a₂+b₂x+c₂y)/2A
N₂=1
N₃
= (a₃+b₃x+c₃y)/2A
N₃=1
N₁ + N₂ + N₃ = 1 (partition of unity) | Nᵢ = 1 at node i, Nᵢ = 0 at other nodes
Fig. 2 — Shape Functions N₁, N₂, N₃: linear interpolation over the triangle
2. Strain-Displacement Matrix [B]
Combining equations (i) and (ii) and differentiating the shape functions gives the $[B]$ matrix ($3 \times 6$) such that:
$$\{\varepsilon\} = [B]\{u\}$$
Full [B] Matrix
$$[B] = \frac{1}{2A} \begin{bmatrix} b_1 & b_2 & b_3 & 0 & 0 & 0 \\ 0 & 0 & 0 & c_1 & c_2 & c_3 \\ c_1 & c_2 & c_3 & b_1 & b_2 & b_3 \end{bmatrix}$$
Expanded using coordinate differences
$$[B] = \frac{1}{2A} \begin{bmatrix} y_{23} & y_{31} & y_{12} & 0 & 0 & 0 \\ 0 & 0 & 0 & x_{32} & x_{13} & x_{21} \\ x_{32} & x_{13} & x_{21} & y_{23} & y_{31} & y_{12} \end{bmatrix}$$
Strain-Displacement Matrix [B] — Structure and Meaning
x-displacements
y-displacements
u₁
u₂
u₃
v₁
v₂
v₃
εx →
1
εy
[B] =
2A
γxy →
b₁
b₂
b₃
0
0
0
← Row 1: Normal strain εx
0
0
0
c₁
c₂
c₃
← Row 2: Normal strain εy
c₁
c₂
c₃
b₁
b₂
b₃
← Row 3: Shear strain γxy
bi = yj − yk | ci = xk − xj | {ε} = [B]{u} | [B] is CONSTANT → strain is constant throughout element
Equivalently: b₁=y₂₃, b₂=y₃₁, b₃=y₁₂ | c₁=x₃₂, c₂=x₁₃, c₃=x₂₁
Fig. 3 — [B] Matrix structure: color-coded by DOF group and strain component
Key Property: All entries of $[B]$ are constants (independent of $x$ and $y$). Therefore the strain is uniform (constant) everywhere within the element — this is why the element is called the Constant Strain Triangle .
3. Stiffness Matrix for CST
3.1 Stress Field on CST Element
Consider a CST element subjected to constant stresses $\sigma_x$, $\sigma_y$, $\tau_{xy}$. Let $t$ be the element thickness.
Stress Field on CST Element — σₓ, σᵧ, τₓᵧ
x
y
1
2
3
σₓ
σₓ
σᵧ
σᵧ
τₓᵧ
τₓᵧ
σₓ — Normal stress (x)
σᵧ — Normal stress (y)
τₓᵧ — Shear stress
Fig. 4 — Stress field on CST element: σₓ (red), σᵧ (green), τₓᵧ (blue)
3.2 Mid-Side Forces
The mid-side forces at each face are computed by integrating the traction over the face:
Force Expression $F_{m1x}$ $\sigma_x \cdot y_{32} \cdot t + \tau_{xy} \cdot x_{23} \cdot t$ $F_{m1y}$ $\sigma_y \cdot x_{23} \cdot t + \tau_{xy} \cdot y_{32} \cdot t$ $F_{m2x}$ $-\sigma_x \cdot y_{31} \cdot t + \tau_{xy} \cdot x_{31} \cdot t$ $F_{m2y}$ $\sigma_y \cdot x_{31} \cdot t - \tau_{xy} \cdot y_{31} \cdot t$ $F_{m3x}$ $\sigma_x \cdot y_{21} \cdot t - \tau_{xy} \cdot x_{21} \cdot t$ $F_{m3y}$ $-\sigma_y \cdot x_{21} \cdot t + \tau_{xy} \cdot y_{21} \cdot t$
3.3 Equivalent Nodal Forces
Each nodal force is the average of the two adjacent mid-side forces (work equivalence):
Force Expression $F_{1x} = \tfrac{1}{2}(F_{m2x}+F_{m3x})$ $\dfrac{t}{2}\bigl(\sigma_x y_{23} + \tau_{xy} x_{32}\bigr)$ $F_{1y} = \tfrac{1}{2}(F_{m2y}+F_{m3y})$ $\dfrac{t}{2}\bigl(\sigma_y x_{32} + \tau_{xy} y_{23}\bigr)$ $F_{2x} = \tfrac{1}{2}(F_{m1x}+F_{m3x})$ $\dfrac{t}{2}\bigl(\sigma_x y_{31} + \tau_{xy} x_{13}\bigr)$ $F_{2y} = \tfrac{1}{2}(F_{m1y}+F_{m3y})$ $\dfrac{t}{2}\bigl(\sigma_y x_{13} + \tau_{xy} y_{31}\bigr)$ $F_{3x} = \tfrac{1}{2}(F_{m1x}+F_{m2x})$ $\dfrac{t}{2}\bigl(\sigma_x y_{12} + \tau_{xy} x_{21}\bigr)$ $F_{3y} = \tfrac{1}{2}(F_{m1y}+F_{m2y})$ $\dfrac{t}{2}\bigl(\sigma_y x_{21} + \tau_{xy} y_{12}\bigr)$
Equivalent Nodal Forces on CST Element — F₁ₓ, F₁ᵧ, F₂ₓ, F₂ᵧ, F₃ₓ, F₃ᵧ
y
1
2
3
F₁ₓ
F₁ᵧ
F₂ₓ
F₂ᵧ
F₃ₓ
F₃ᵧ
F₁ₓ = (t/2){σₓy₂₃ + τₓᵧx₃₂}
F₁ᵧ = (t/2){σᵧx₃₂ + τₓᵧy₂₃}
F₂ₓ = (t/2){σₓy₃₁ + τₓᵧx₁₃}
F₂ᵧ = (t/2){σᵧx₁₃ + τₓᵧy₃₁}
F₃ₓ = (t/2){σₓy₁₂ + τₓᵧx₂₁}
F₃ᵧ = (t/2){σᵧx₂₁ + τₓᵧy₁₂}
{F} = (t/2)·[B]ᵀ·{σ}·2A ∴ {F} = t·A·[B]ᵀ[D][B]·{u} = [K]{u}
Fig. 5 — Equivalent nodal forces on CST element
3.4 Force Vector in Matrix Form
$$\{F\} = \frac{t}{2} \cdot [B]^T \cdot \{\sigma\} \times 2A$$
Using the stress-strain relation $\{\sigma\} = [D]\{\varepsilon\}$ and $\{\varepsilon\} = [B]\{u\}$:
$$\{F\} = \frac{t}{2} \times 2A \cdot [B]^T [D] [B]\{u\} = t \cdot A \cdot [B]^T [D] [B]\{u\}$$
3.5 Stiffness Matrix
Comparing with $\{F\} = [K]\{u\}$, we get the $6 \times 6$ element stiffness matrix:
$$\boxed{[K] = t \cdot A \cdot [B]^T [D] [B]}$$
Note: Since $[B]$ is a constant matrix, $[K] = tA[B]^T[D][B]$ is simply a matrix product — no numerical integration is needed , making CST computationally very efficient.
4. Forces in CST Element
Forces on CST Element — (i) Body Force (ii) Traction Force
Body Force (Gravity)
fb₃
fb₁
fb₂
fᵦ = γ·A·t / 3 per node
{fᵦ} = (γ·A·t/3)×{0,0,0,−1,−1,−1}ᵀ
Traction Force
i (Pᵢ)
j (Pⱼ)
(traction)
θ
{fᵢₓ, fᵢᵧ, fⱼₓ, fⱼᵧ} = (t·lᵢⱼ/6) ×
{(2Pᵢ+Pⱼ)cosθ, (Pᵢ+2Pⱼ)cosθ,
(2Pᵢ+Pⱼ)sinθ, (Pᵢ+2Pⱼ)sinθ}
Fig. 6 — (Left) Body force distributed to 3 nodes; (Right) Traction force on edge
4.1 Body Force
For a CST element with material unit weight $\gamma$, thickness $t$, and area $A$:
Body Force Calculation
$$\text{Total weight: } W = \gamma \cdot A \cdot t$$
$$\text{Body force per node: } f_b = \frac{W}{3} = \frac{\gamma \cdot A \cdot t}{3}$$
Body force vector (gravity in $-y$ direction):
$$\{f_b\} = \frac{\gamma A t}{3} \begin{Bmatrix} 0 \\ 0 \\ 0 \\ -1 \\ -1 \\ -1 \end{Bmatrix}$$
4.2 Traction Force
For a uniformly varying traction on edge $i$–$j$ (length $l_{ij}$) with intensities $P_i$ at node $i$ and $P_j$ at node $j$, at angle $\theta$ to $x$-axis:
$$\begin{Bmatrix} f_{ix} \\ f_{iy} \\ f_{jx} \\ f_{jy} \end{Bmatrix} = \frac{t \cdot l_{ij}}{6} \begin{Bmatrix} (2P_i + P_j)\cos\theta \\ (2P_i + P_j)\sin\theta \\ (P_i + 2P_j)\cos\theta \\ (P_i + 2P_j)\sin\theta \end{Bmatrix}$$
For uniform traction $P_i = P_j = P$: each loaded node receives $\dfrac{t \cdot l_{ij} \cdot P}{2}$ in each direction.
CST Stiffness Matrix Derivation — Step-by-Step Flowchart
1
Define Node Coordinates
(x₁,y₁), (x₂,y₂), (x₃,y₃)
2
Compute 2A = determinant
|1 x₁ y₁; 1 x₂ y₂; 1 x₃ y₃|
3
Compute b, c coefficients
b₁=y₂₃, b₂=y₃₁, b₃=y₁₂ | c₁=x₃₂, c₂=x₁₃, c₃=x₂₁
4
Assemble [B] = (1/2A)×[…]
3×6 strain-displacement matrix
5
Compute [D] matrix
(plane stress: E/(1−ν²)×[…])
6
[K] = t·A·[B]ᵀ[D][B]
6×6 stiffness matrix
7
Apply BCs, solve {u} = [K]⁻¹{F}
Get nodal displacements
8
{ε}=[B]{u} {σ}=[D][B]{u}
Compute element strains & stresses
Fig. 7 — Flowchart: Step-by-step procedure to compute CST stiffness matrix and stresses
5. Numerical Example 1 — Thin Plate Under Point Loads
Example 1: Thin Plate with Nodal Point Loads
A thin plate is subjected to loads as shown. Thickness $t = 0.3$ inch, Poisson's ratio $\nu = 0.3$, $E = 30 \times 10^6$ psi. Determine the nodal displacements and element stresses.
Numerical Example 1 — Thin Plate with Nodal Point Loads
x
y
1
Fixed
(0,10)
2
Fixed
(0,0)
3
(20,10)
2000 lb
−2000 lb
20 in
10 in
Given:
t = 0.3 inch
ν = 0.3
E = 30×10⁶ psi
Node 1: (0, 10) — fixed
Node 2: (0, 0) — fixed
Node 3: (20, 10) — free
2A = 200 in² ∴ A = 100 in²
Fig. 8 — Example 1: Plate geometry, boundary conditions, and applied loads
Solution
Step 1: Given Data
Parameter Value Thickness $t$ 0.3 inch Poisson's ratio $\nu$ 0.3 Modulus $E$ $30 \times 10^6$ psi Node 1 $(0, 10)$ — fixed Node 2 $(0, 0)$ — fixed Node 3 $(20, 10)$ — free Load at Node 3 $F_x = +2000$ lb, $F_y = -2000$ lb
Step 2: Computation of Area
$$2A = \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \end{vmatrix} = \begin{vmatrix} 1 & 0 & 10 \\ 1 & 0 & 0 \\ 1 & 20 & 10 \end{vmatrix} = 200$$
$$\therefore A = 100 \text{ in}^2$$
Step 3: [B] Matrix Computation
Coordinate differences:
$y_{23} = 0 - 10 = -10$ $y_{31} = 10 - 10 = 0$ $y_{12} = 10 - 0 = 10$ $x_{32} = 20 - 0 = 20$ $x_{13} = 0 - 20 = -20$ $x_{21} = 0 - 0 = 0$
$$[B] = \frac{1}{200} \begin{bmatrix} -10 & 0 & 10 & 0 & 0 & 0 \\ 0 & 0 & 0 & 20 & -20 & 0 \\ 20 & -20 & 0 & -10 & 0 & 10 \end{bmatrix}$$
Simplified ($B = \frac{1}{20}[\dots]$):
$$[B] = \frac{1}{20} \begin{bmatrix} -1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & -2 & 0 \\ 2 & -2 & 0 & -1 & 0 & 1 \end{bmatrix}$$
Step 4: Elasticity Matrix [D] (Plane Stress)
Elasticity Matrix [D] — Plane Stress Formulation
General $[D]$ =
E
1−ν²
×
1
ν
0
ν
1
0
0
0
(1−ν)/2
← coupling between εx & εy
← symmetric
← shear: (1−ν)/2
For ν=0.3, E=30×10⁶ psi → [D] = 32.967×10⁶ ×
1
0.3
0
0.3
1
0
0
0
0.35
Fig. 9 — [D] Matrix for plane stress: general form and evaluated for ν = 0.3
$$[D] = \frac{E}{1-\nu^2} \begin{bmatrix} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac{1-\nu}{2} \end{bmatrix} = \frac{30 \times 10^6}{1 - 0.09} \begin{bmatrix} 1 & 0.3 & 0 \\ 0.3 & 1 & 0 \\ 0 & 0 & 0.35 \end{bmatrix}$$
$$= 32.967 \times 10^6 \begin{bmatrix} 1 & 0.3 & 0 \\ 0.3 & 1 & 0 \\ 0 & 0 & 0.35 \end{bmatrix}$$
Step 5: Stiffness Matrix [K] = tA[B]ᵀ[D][B]
After matrix multiplication:
$$[K] = 24{,}725{,}274.73 \times \begin{bmatrix} 2.4 & -1.4 & -1 & -1.3 & 0.6 & 0.7 \\ -1.4 & 1.4 & 0 & 0.7 & 0 & -0.7 \\ -1 & 0 & 1 & 0.6 & -0.6 & 0 \\ -1.3 & 0.7 & 0.6 & 4.35 & -4 & -0.35 \\ 0.6 & 0 & -0.6 & -4 & 4 & 0 \\ 0.7 & -0.7 & 0 & -0.35 & 0 & 0.35 \end{bmatrix}$$
Step 6: Applying Boundary Conditions
Nodes 1 and 2 are fixed: $u_1 = u_2 = v_1 = v_2 = 0$. Active DOFs: $u_3$, $v_3$.
Reduced system:
$$\begin{Bmatrix} 2000 \\ -2000 \end{Bmatrix} = 24{,}725{,}274.73 \times \begin{bmatrix} 1 & 0 \\ 0 & 0.35 \end{bmatrix} \begin{Bmatrix} u_3 \\ v_3 \end{Bmatrix}$$
Step 7: Displacement Solution
$$\begin{Bmatrix} u_3 \\ v_3 \end{Bmatrix} = \frac{1}{24{,}725{,}274.73} \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{0.35} \end{bmatrix} \begin{Bmatrix} 2000 \\ -2000 \end{Bmatrix}$$
$$\boxed{u_3 = 8.088 \times 10^{-5} \text{ inch}, \quad v_3 = -2.311 \times 10^{-4} \text{ inch}}$$
Step 8: Element Stresses
With $\{u\} = \{0,\; 0,\; 8.088 \times 10^{-5},\; 0,\; 0,\; -2.311 \times 10^{-4}\}^T$:
$$\{\sigma\} = [D][B]\{u\}$$
$$\boxed{\sigma_x = 133.333 \text{ lb/in}^2, \quad \sigma_y = 40.000 \text{ lb/in}^2, \quad \tau_{xy} = -133.333 \text{ lb/in}^2}$$
6. Numerical Example 2 — Steel Plate with Body & Traction Forces
Example 2: Equilateral Triangular Steel Plate
Determine the stresses and strains at the CG of a steel plate of thickness 10 mm. $E = 210 \times 10^3$ MPa, $\nu = 0.3$, $\gamma = 78.5$ kN/m³, side = 100 mm (equilateral triangle). Traction $T = 10$ kN/m² at 30° on face 2-3.
Numerical Example 2 — Equilateral Triangular Steel Plate
x
y
1
(−0.05, 0)
2
(0.05, 0)
3
(0, 0.0866)
γ=78.5 kN/m³
(self-weight)
T=10 kN/m²
θ=30°
100 mm
Given:
E = 210×10³ MPa
ν = 0.3
t = 10 mm = 0.01 m
γ = 78.5 kN/m³
Side = 100 mm
T = 10 kN/m² @ 30°
Nodes 1,2: pinned
A = 0.00433 m²
2A = 0.00866 m²
Fig. 10 — Example 2: Equilateral triangular steel plate with body force and traction load
Solution
Step 1: Given Data
Parameter Value $E$ $210 \times 10^3$ MPa $= 210 \times 10^6$ kN/m² $\nu$ 0.3 $t$ 10 mm = 0.01 m $\gamma$ 78.5 kN/m³ Side length 100 mm = 0.1 m Node 1 $(-0.05,\; 0)$ — pinned Node 2 $(0.05,\; 0)$ — pinned Node 3 $(0,\; 0.0866)$ — free Traction $T$ 10 kN/m² at $\theta = 30°$ on face 2-3
Step 2: Area
$$2A = \begin{vmatrix} 1 & -0.05 & 0 \\ 1 & 0.05 & 0 \\ 1 & 0 & 0.0866 \end{vmatrix} = 0.00866 \text{ m}^2 \quad \therefore A = 0.00433 \text{ m}^2$$
Step 3: Force Vector — Body Force
Gravity distributed equally to 3 nodes:
$$\{f_b\} = \frac{\gamma \cdot A \cdot t}{3} \times \{-1,\; -1,\; -1\}^T$$
$$= \frac{78.5 \times 0.00433 \times 0.01}{3} \times \{-1,\; -1,\; -1\}^T$$
$$\{f_b\} = \{-0.001133,\; -0.001133,\; -0.001133\}^T \text{ kN (y-components)}$$
Step 4: Force Vector — Traction Force
For uniform traction $T = 10$ kN/m² on face 2-3, length = 0.1 m, $\theta = 30°$:
$$\{f^t\} = \frac{t \cdot l}{2} \times \{T\cos\theta,\; T\sin\theta,\; T\cos\theta,\; T\sin\theta\}^T$$
$$= \frac{0.01 \times 0.1}{2} \times \{10\cos 30°,\; 10\sin 30°,\; 10\cos 30°,\; 10\sin 30°\}^T$$
$$\{f^t\} = \{0.00433,\; 0.0025,\; 0.00433,\; 0.0025\}^T \text{ kN (nodes 2, 3: x, y)}$$
Combined force vector $\{F\}$:
$F_{1x} = R_{1x}$ $F_{1y} = R_{1y} - 0.001133$ $F_{2x} = R_{2x} + 0.00433$ $F_{2y} = R_{2y} + 0.001367$ $F_{3x} = 0.00433$ $F_{3y} = 0.001367$
Step 5: [B] Matrix
$$[B] = \frac{1}{2A} \begin{bmatrix} -10 & 0 & 10 & 0 & 0 & 0 \\ 0 & -5.774 & 0 & -5.774 & 0 & 11.549 \\ -5.774 & -10 & -5.774 & 10 & 11.549 & 0 \end{bmatrix}$$
Step 6: [D] Matrix
$$[D] = \frac{210 \times 10^6}{1 - 0.09} \begin{bmatrix} 1 & 0.3 & 0 \\ 0.3 & 1 & 0 \\ 0 & 0 & 0.35 \end{bmatrix} = 2.308 \times 10^8 \begin{bmatrix} 1 & 0.3 & 0 \\ 0.3 & 1 & 0 \\ 0 & 0 & 0.35 \end{bmatrix}$$
Step 7: Stiffness Matrix [K]
After computing $K = tA[B]^T[D][B]$:
$$\frac{[K]}{9.99 \times 10^3} = \begin{bmatrix} 111.667 & 37.528 & -88.333 & -2.887 & -23.333 & -34.641 \\ 37.528 & 68.333 & 2.887 & -1.667 & -40.415 & -66.667 \\ -88.333 & 2.887 & 111.667 & -37.528 & -23.333 & 34.641 \\ -2.887 & -1.667 & -37.528 & 68.333 & 40.415 & -66.667 \\ -23.333 & -40.415 & -23.333 & 40.415 & 46.667 & 0 \\ -34.641 & -66.667 & 34.641 & -66.667 & 0 & 133.333 \end{bmatrix}$$
Step 8: Displacement Solution
Applying BCs ($u_1 = v_1 = u_2 = v_2 = 0$), reduced system:
$$\begin{Bmatrix} 0.00433 \\ 0.001367 \end{Bmatrix} = 9.99 \times 10^3 \begin{bmatrix} 46.667 & 0 \\ 0 & 133.333 \end{bmatrix} \begin{Bmatrix} u_3 \\ v_3 \end{Bmatrix}$$
$$\boxed{u_3 = 9.29 \times 10^{-9} \text{ m}, \quad v_3 = 1.03 \times 10^{-9} \text{ m}}$$
Step 9: Element Stresses
$\{\sigma\} = [D][B]\{u\}$:
$$\boxed{\sigma_x = 0.82 \text{ kN/m}^2, \quad \sigma_y = 2.734 \text{ kN/m}^2, \quad \tau_{xy} = 8.66 \text{ kN/m}^2}$$
Step 10: Element Strains
$\{\varepsilon\} = [B]\{u\}$:
$$\boxed{\varepsilon_x = 0, \quad \varepsilon_y = 1.18 \times 10^{-8}, \quad \gamma_{xy} = 1.07 \times 10^{-7}}$$
7. Summary — CST Element Properties
Property Detail Element Type 2D triangular, plane stress / plane strain Nodes 3 corner nodes DOF per node 2 ($u$, $v$) Total DOF 6 Shape function order Linear (bilinear in space) Strain field Constant throughout element Stress field Constant throughout element $[B]$ matrix size $3 \times 6$ (constant matrix) $[D]$ matrix size $3 \times 3$ $[K]$ matrix size $6 \times 6$ Stiffness formula $[K] = t \cdot A \cdot [B]^T [D] [B]$ Element stress $\{\sigma\} = [D][B]\{u\}$ Element strain $\{\varepsilon\} = [B]\{u\}$ Body force (per node) $f_b = \gamma \cdot A \cdot t\, /\, 3$ Traction force See Section 4.2 formula Key advantage Simple, constant $[B]$, no integration needed Key limitation Poor accuracy for high stress gradients Best used for Irregular geometries, mesh transitions