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Finite Element Methods · Chapter 6 of 10 · Advanced

Plane Stress and the Constant-Strain Triangle

Two dimensions bring real geometry into reach. The simplest 2D element, a triangle with a single constant strain, shows the whole machinery: a B matrix from geometry, a D matrix from the material, and a stiffness from their product.

Readiness check

This chapter moves to two-dimensional elasticity. Tick only what you can do closed-notes.

Recall plane stress and Hooke's law with Poisson's ratio.
Find the area of a triangle from coordinates.
Recall strain as a derivative of displacement.
Multiply a 3-by-3 matrix by a vector.
Distinguish a thin plate from a thick body.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRevisit the strain-displacement matrix in Chapter 5.

3 or more weak itemsReview stress and strain in Mechanics of Materials, Chapter 1.

The core idea

A three-node triangle interpolates two displacements per node, giving a single constant strain throughout. Its stiffness is t·A·BᵀDB, where B comes from the geometry and D from the material.

CST: 3 nodes × 2 DOF = 6 DOFk = t A BᵀDB (constant B)plane stress: D = E/(1−ν²)[...]

Real parts are two-dimensional, so the field is now two displacements, u and v, at each node. The constant-strain triangle has three nodes and six degrees of freedom; because its shape functions are linear, the strain it represents is constant across the element, which is why it is called the constant-strain triangle. The strain-displacement matrix B is built from the nodal coordinates and the element area, and being constant, it factors out of the stiffness integral, leaving the compact stiffness k = t·A·BᵀDB for a plate of thickness t and area A. The material enters through the constitutive matrix D: plane stress (σ_z = 0) for thin plates, plane strain (ε_z = 0) for long bodies. Many small triangles approximate a curved, loaded part.

The skill works when: you build B from the geometry, choose plane stress or strain for D, and form t·A·BᵀDB.

The skill breaks down when: a single constant-strain triangle is trusted where strain actually varies steeply.

The concept. The constant-strain triangle: three nodes, each with two displacement degrees of freedom (u, v). Its linear field gives a single constant strain throughout the element.

The skills, taught in order

Five skills set up 2D elasticity, the CST element, the B matrix, the D matrix, and the element stiffness.

6.1 Plane stress and plane strain

Two idealisations reduce three-dimensional elasticity to two. Plane stress assumes no through-thickness stress (σ_z = 0), valid for thin plates loaded in their plane. Plane strain assumes no through-thickness strain (ε_z = 0), valid for long bodies like dams. Each has its own constitutive matrix.

Idealisation	Assumption	Typical use
Plane stress	σ_z = 0	thin plates and brackets
Plane strain	ε_z = 0	dams, tunnels, long bodies

6.2 The constant-strain triangle

The CST has three nodes and six degrees of freedom, with linear shape functions in two coordinates. Because the field is linear, its derivatives, the strains, are constant over the element, so each triangle carries a single uniform strain and stress state.

6.3 The B matrix from geometry

The strain-displacement matrix B is assembled from the nodal coordinates: its entries are differences of coordinates divided by twice the element area. The area itself is A = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. B is constant for a CST.

6.4 The D matrix from the material

For plane stress, D = E/(1 − ν²) times the matrix [1, ν, 0; ν, 1, 0; 0, 0, (1 − ν)/2], relating the stress vector to the strain vector. It encodes how Poisson coupling links the in-plane stresses.

6.5 The element stiffness

Because B is constant, the stiffness integral collapses to k = t·A·BᵀDB, with t the thickness and A the area. Assembling these triangle stiffnesses and solving gives nodal displacements, from which the constant element stresses are recovered.

Engineering connection: the CST's constant strain is its weakness; isoparametric quadrilaterals in Chapter 7 capture varying strain far more accurately.

Worked example 1: the area of a triangle element

A constant-strain triangle has nodes at (0, 0), (2, 0), and (0, 1) in millimetres. Find its area, and one entry of the strain-displacement relation (the coordinate difference b₁ = y₂ − y₃).

Figure 1. A right-triangle element with legs 2 and 1. Its area enters the stiffness and the strain-displacement matrix, both of which are constant for a CST.

ProblemFind the area and the entry b₁ for the triangle in Figure 1.
Given / findNodes (0, 0), (2, 0), (0, 1). Find A and b₁ = y₂ − y₃.
AssumptionsConstant-strain triangle; coordinates in millimetres.
ModelA = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|; b₁ = y₂ − y₃.
EquationsA = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|b₁ = y₂ − y₃
SolveA = ½|0(0 − 1) + 2(1 − 0) + 0(0 − 0)| = ½|2| = 1 mm². b₁ = y₂ − y₃ = 0 − 1 = −1.
CheckThe shoelace area equals ½ × base × height = ½ × 2 × 1 = 1, confirming the formula. The b coefficients (differences of y) and c coefficients (differences of x) build the constant B matrix.
ConclusionThe geometry alone fixes the area and the B matrix for a CST. Both are constant, which is what makes the element so simple and so limited.

Result. A = 1 mm²; b₁ = −1.

Worked example 2: stress from the plane-stress matrix

A thin steel plate (E = 200 GPa, ν = 0.3) is in plane stress. An element has the strain state ε_x = 0.001, ε_y = 0, γ_xy = 0. Find the in-plane stresses using the D matrix.

Figure 2. The plane-stress D matrix couples the strains: a strain only in x still produces a stress in y through Poisson's ratio, here about a third of σ_x.

ProblemFind σ_x and σ_y for the strain state in Figure 2.
Given / findE = 200 000 MPa, ν = 0.3, ε_x = 0.001, ε_y = 0, γ_xy = 0. Find σ_x, σ_y.
AssumptionsPlane stress; linear isotropic material.
Modelσ = Dε with D = E/(1 − ν²) times the plane-stress matrix.
EquationsE/(1 − ν²) = 200 000/0.91 = 219 780 MPaσ_x = (E/(1−ν²))(ε_x + νε_y), σ_y = (E/(1−ν²))(νε_x + ε_y)
Solveσ_x = 219 780 × (0.001 + 0) = 219.8 MPa. σ_y = 219 780 × (0.3 × 0.001) = 65.9 MPa. τ_xy = 0.
CheckThe y-stress is ν times the x-stress, 0.3 × 219.8 = 65.9 MPa, the Poisson coupling that holds ε_y at zero. A pure uniaxial stress would instead let the element contract in y.
ConclusionThe D matrix converts the element's constant strain into its constant stress. Plane stress and plane strain differ only in this matrix.

Result. σ_x = 219.8 MPa, σ_y = 65.9 MPa, τ_xy = 0.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Wrong idealisation	Plane stress used on a thick body	"Thin plate or long body?"	Plane stress for thin, plane strain for long; pick the right D.
One CST where strain varies	Stress badly under-predicted	"Is the strain really constant here?"	Refine the mesh or use a higher-order element.
Forgetting thickness	Stiffness off by the factor t	"Did I include the plate thickness?"	The stiffness is t·A·BᵀDB.
Ignoring Poisson coupling	Missing the transverse stress	"Did I use the full D matrix?"	The off-diagonal ν terms couple the in-plane stresses.

Practice ladder

Level 1 · Direct skill

Find the area of a triangle with nodes (0, 0), (4, 0), (0, 3).

Show answer

A = ½|0(0 − 3) + 4(3 − 0) + 0(0 − 0)| = ½ × 12 = 6 area units.

Level 2 · Mixed concept

For the Worked Example 2 material, find σ_x if the strain is doubled to ε_x = 0.002 (ε_y = 0).

Show answer

σ_x = 219 780 × 0.002 = 439.6 MPa, double the previous value; σ_y = 131.9 MPa.

Level 3 · Independent problem

Find E/(1 − ν²) for aluminium (E = 70 GPa, ν = 0.33), and the resulting σ_x for ε_x = 0.001, ε_y = 0.

Show answer

E/(1 − ν²) = 70 000/(1 − 0.1089) = 70 000/0.8911 = 78 555 MPa. σ_x = 78 555 × 0.001 = 78.6 MPa.

Level 4 · Transfer to real engineering

For a real thin part, decide whether plane stress or plane strain applies, and describe how you would mesh a region of high stress gradient.

What good work looks like

A correct idealisation from the geometry, and a refined mesh (or higher-order elements) where the strain varies steeply, since the CST is constant-strain.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I chose plane stress or plane strain correctly for this part."

"Give me three triangles; I will compute their areas."

"Build the CST stiffness for me." Forming B and D yourself is the skill.

"What are the stresses?" Multiplying Dε yourself is the point.

Portfolio task

For a thin loaded part, set up a CST: compute the element area, choose the D matrix, form B from the geometry, and recover the constant element stress.

Must include: an element area, a chosen D matrix, and a recovered element stress.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Plane stress versus plane strain?

Plane stress sets σ_z = 0 (thin plate); plane strain sets ε_z = 0 (long body).

2. Why is the triangle called constant-strain?

Its linear field gives constant derivatives, so the strain is uniform over the element.

3. How is the element area found?

A = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

4. Write the plane-stress D factor.

E/(1 − ν²) times the matrix [1, ν, 0; ν, 1, 0; 0, 0, (1 − ν)/2].

5. Write the CST stiffness.

k = t·A·BᵀDB, with B and A constant.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the area and the stress from a blank page.

+3 daysSet up two new CST elements.

+7 daysCarry the B and D matrices into isoparametric elements, Chapter 7.

+30 daysReuse plane stress when validating a 2D FEA.

Textbook mapping

Item	Mapping
Primary source	Logan, A First Course in the Finite Element Method, Chapter 6 (Plane Stress and Plane Strain)
Cross-reference	Hutton, Ch. 6 · Mechanics of Materials, Ch. 1
Core topics	6.1 Plane stress and strain · 6.2 The CST · 6.3 B matrix · 6.4 D matrix · 6.5 Element stiffness
Engineering connection	Isoparametric quadrilaterals overcome the CST's constant-strain limitation.
Read next	Chapter 7: Isoparametric Formulation and Gauss Quadrature.