Constitutive Relations for 2D Analysis

TL;DR

Constitutive relations describe how materials deform under stress, linking stress and strain. For 2D analysis, we simplify these relationships based on plane stress or plane strain assumptions. These relations are crucial for predicting a structure's behavior in CAE simulations.

1. The Mental Model

Imagine you're squishing or stretching a piece of play-doh. How much it deforms depends on how hard you push (stress) and how stiff the play-doh is (material properties). Constitutive relations are just the mathematical rules that describe this squishing and stretching for different materials.

2. The Core Material

In Computer Aided Engineering (CAE), especially when you're analyzing structures, you need to understand how materials respond to applied loads. This response is governed by constitutive relations, which are mathematical equations connecting stress (force per unit area) and strain (deformation per unit length).

For 2D analysis, things simplify because we make assumptions about the stress or strain state. The two main types are plane stress and plane strain.

Plane Stress

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You use plane stress when you have a thin plate or shell loaded in its own plane. Think of a thin metal sheet with forces applied only along its edges. In this scenario, the stress perpendicular to the plane of the sheet (the z-direction) is assumed to be zero.

Mathematically, this means:
$\sigma_z = 0$, $\tau_{xz} = 0$, $\tau_{yz} = 0$

The remaining stress components are $\sigma_x, \sigma_y, \tau_{xy}$.

The constitutive relation for an isotropic (properties are the same in all directions) linear elastic material under plane stress is:

$$ \begin{bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{bmatrix} = \frac{1}{E} \begin{bmatrix} 1 & -\nu & 0 \\ -\nu & 1 & 0 \\ 0 & 0 & 2(1+\nu) \end{bmatrix} \begin{bmatrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{bmatrix} $$

Where:
* $E$ is Young's Modulus (a measure of stiffness).
* $\nu$ is Poisson's Ratio (describes transverse contraction).
* $\varepsilon_x, \varepsilon_y$ are normal strains.
* $\gamma_{xy}$ is shear strain.

You can also write this in terms of stress from strain:

$$ \begin{bmatrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{bmatrix} = \frac{E}{1-\nu^2} \begin{bmatrix} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac{1-\nu}{2} \end{bmatrix} \begin{bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{bmatrix} $$

Plane Strain

You apply plane strain when you have a very thick structure where the deformation in one direction (usually the z-direction) is negligible. Imagine a long dam or a thick pipe where the length is much greater than the cross-section, and the loads don't vary along the length. Here, the strain perpendicular to the plane of interest is assumed to be zero.

Mathematically, this means:
$\varepsilon_z = 0$, $\gamma_{xz} = 0$, $\gamma_{yz} = 0$

The remaining strain components are $\varepsilon_x, \varepsilon_y, \gamma_{xy}$.

The constitutive relation for an isotropic linear elastic material under plane strain is:

$$ \begin{bmatrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{bmatrix} = \frac{E}{(1+\nu)(1-2\nu)} \begin{bmatrix} 1-\nu & \nu & 0 \\ \nu & 1-\nu & 0 \\ 0 & 0 & \frac{1-2\nu}{2} \end{bmatrix} \begin{bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{bmatrix} $$

Notice that even though $\varepsilon_z=0$, the stress $\sigma_z$ is generally not zero. It can be calculated as $\sigma_z = \nu(\sigma_x + \sigma_y)$.

When to Use Which

Choosing the right assumption is critical for accurate simulations. Here's a quick guide:

graph TD
    A["Structure Type & Loading"] --> B{Is it thin & loaded in-plane?};
    B -- Yes --> C["Assume Plane Stress"];
    B -- No --> D{Is it thick & deformation constrained out-of-plane?};
    D -- Yes --> E["Assume Plane Strain"];
    D -- No --> F["Requires Full 3D Analysis"];
    C --> G["Use Plane Stress Constitutive Matrix"];
    E --> G;
    F --> H["Use 3D Constitutive Matrix"];

Isotropic vs. Anisotropic Materials

The equations above are for isotropic materials (e.g., most metals). If your material is anisotropic (properties vary with direction, like wood or composites), the constitutive matrices will be more complex and have more non-zero terms. For 2D, this often involves orthotropic materials, where properties are different along two or three orthogonal axes.

3. Worked Example

Let's consider a thin aluminum plate under a 2D stress state.
Given:
* Young's Modulus (E) = 70 GPa = $70 \times 10^9$ Pa
* Poisson's Ratio ($\nu$) = 0.33
* Applied stresses: $\sigma_x = 100$ MPa, $\sigma_y = 50$ MPa, $\tau_{xy} = 20$ MPa

Since it's a thin plate, we'll assume plane stress. We want to find the resulting strains ($\varepsilon_x, \varepsilon_y, \gamma_{xy}$).

Using the plane stress constitutive relation (stress to strain form):
$$ \begin{bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{bmatrix} = \frac{1}{E} \begin{bmatrix} 1 & -\nu & 0 \\ -\nu & 1 & 0 \\ 0 & 0 & 2(1+\nu) \end{bmatrix} \begin{bmatrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{bmatrix} $$

First, let's calculate $2(1+\nu) = 2(1+0.33) = 2(1.33) = 2.66$.

Now, plug in the values:
$$ \begin{bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{bmatrix} = \frac{1}{70 \times 10^9} \begin{bmatrix} 1 & -0.33 & 0 \\ -0.33 & 1 & 0 \\ 0 & 0 & 2.66 \end{bmatrix} \begin{bmatrix} 100 \times 10^6 \\ 50 \times 10^6 \\ 20 \times 10^6 \end{bmatrix} $$

Calculate each strain component:
$\varepsilon_x = \frac{1}{70 \times 10^9} (1 \times 100 \times 10^6 + (-0.33) \times 50 \times 10^6 + 0 \times 20 \times 10^6)$
$\varepsilon_x = \frac{1}{70 \times 10^9} (100 \times 10^6 - 16.5 \times 10^6) = \frac{83.5 \times 10^6}{70 \times 10^9} \approx 0.001193$

$\varepsilon_y = \frac{1}{70 \times 10^9} (-0.33 \times 100 \times 10^6 + 1 \times 50 \times 10^6 + 0 \times 20 \times 10^6)$
$\varepsilon_y = \frac{1}{70 \times 10^9} (-33 \times 10^6 + 50 \times 10^6) = \frac{17 \times 10^6}{70 \times 10^9} \approx 0.000243$

$\gamma_{xy} = \frac{1}{70 \times 10^9} (0 \times 100 \times 10^6 + 0 \times 50 \times 10^6 + 2.66 \times 20 \times 10^6)$
$\gamma_{xy} = \frac{1}{70 \times 10^9} (53.2 \times 10^6) \approx 0.000760$

So, the strains are approximately:
$\varepsilon_x = 0.001193$
$\varepsilon_y = 0.000243$
$\gamma_{xy} = 0.000760$

These are dimensionless values, representing deformation per unit length.

4. Key Takeaways

• Constitutive relations mathematically link stress and strain, forming the backbone of material behavior in CAE.
• For 2D analysis, you primarily use either plane stress or plane strain assumptions, simplifying calculations.
• Plane stress assumes zero stress perpendicular to a thin plate, suitable for thin structures loaded in their plane.
• Plane strain assumes zero strain perpendicular to a thick body, applicable to long structures with uniform cross-section.
• The choice between plane stress and plane strain drastically affects the calculated stresses and strains.
• Material properties like Young's Modulus and Poisson's Ratio are essential inputs to these relations.

Common Mistakes to Avoid:
- Mixing up plane stress and plane strain: These are distinct assumptions with different constitutive matrices.
- Incorrectly applying units: Always ensure consistent units (e.g., Pa for stress, dimensionless for strain).
- Assuming zero out-of-plane stress for plane strain: While out-of-plane strain is zero, out-of-plane stress ($\sigma_z$) is generally not zero in plane strain.
- Using isotropic relations for anisotropic materials: Real-world composites require more complex constitutive models.

5. Now Try It

You're analyzing a very thick concrete wall, effectively under plane strain conditions. The material properties are: E = 30 GPa, $\nu$ = 0.2. If the measured strains are $\varepsilon_x = 0.000$005$, $\varepsilon_y = -0.0001$, and $\gamma_{xy} = 0.0002$, calculate the corresponding stresses ($\sigma_x, \sigma_y, \tau_{xy}$). What does a negative $\varepsilon_y