Assumptions, Rigidity, and Governing Equations

In solid mechanics, another reduction to 2D is plane stain. It is used to model bodies that are very long in one direction (the axial direction) compared to their other two dimensions, with loads and geometry that do not vary along this long axis. A typical example is a long dam, a tunnel, or a retaining wall under uniform loading.

Let the cross-section of the structure be the \(xy\)-plane, and the long (axial) direction be the \(z\)-axis.

1. Classical Plane Strain Formulation

Primary Assumptions: The core assumption of plane strain is kinematic (related to deformation). For a very long body constrained at its ends and loaded uniformly along its length, it is assumed that every cross-section deforms identically and there is no displacement in the long axial direction.

Mathematically, this dictates: \[ w(x,y,z) = 0 \] \[ u = u(x,y), \quad v = v(x,y) \] The in-plane displacements, u and v, are functions of x and y only.

From these kinematic assumptions, three strain components are immediately zero: \[ \epsilon_{zz} = \frac{\partial w}{\partial z} = 0 \] \[ \gamma_{xz} = \frac{\partial u}{\partial z} + \frac{\partial w}{\partial x} = 0 \] \[ \gamma_{yz} = \frac{\partial v}{\partial z} + \frac{\partial w}{\partial y} = 0 \]

The Axial Stress (σ_zz): A crucial consequence of the plane strain condition is that even though the axial strain (\(\epsilon_{zz}\)) is zero, the axial stress (\(\sigma_{zz}\)) is not zero. The material’s tendency to contract or expand in the z-direction due to the Poisson effect from in-plane stresses is physically constrained. This constraint generates a reaction stress, \(\sigma_{zz}\).

From the generalized Hooke’s Law for \(\epsilon_{zz}\): \[ \epsilon_{zz} = \frac{1}{E} [\sigma_{zz} - \nu(\sigma_{xx} + \sigma_{yy})] = 0 \] Solving for \(\sigma_{zz}\) gives: \[ \sigma_{zz} = \nu(\sigma_{xx} + \sigma_{yy}) \] This shows that an axial stress develops to enforce the zero axial strain condition.

Mathematical Consistency: Unlike plane stress, the classical plane strain formulation is kinematically consistent. The assumptions about displacement lead directly to the defined strain state without creating any internal contradictions with the equations of 3D elasticity. The formulation is not an approximation in the same way as plane stress; rather, it represents an exact solution for an idealized physical situation (an infinitely long body).

2. Generalized Plane Strain

While classical plane strain is mathematically consistent, its assumption of w = 0 (and thus ε_zz = 0) is highly restrictive. It cannot, for example, model a long cylinder with free ends that is subjected to a uniform temperature change, as the body would need to expand or contract along the z-axis.

Generalized plane strain is an extension that relaxes this strict constraint. It allows for a uniform axial extension and/or bending of the body. The simplest form assumes that the axial strain is a constant: \[ \epsilon_{zz} = \epsilon_0 \quad (\text{a constant}) \] This corresponds to a displacement field where u and v are still only functions of x and y, but the axial displacement w is allowed to be a linear function of z. This formulation can handle problems involving net axial forces or uniform thermal expansion while still keeping the problem’s governing equations two-dimensional.

3. Summary of Equations and Unknowns (Classical Case)

For the classical plane strain problem (\(\epsilon_{zz}=0\)), the system is mathematically determinate.

The Unknowns (Total: 8): * Displacements (2): \(u(x,y), v(x,y)\) * Strains (3): \(\epsilon_{xx}, \epsilon_{yy}, \gamma_{xy}\) * Stresses (3): \(\sigma_{xx}, \sigma_{yy}, \sigma_{xy}\) (Note: \(\sigma_{zz}\) is also an unknown but is directly determined by \(\sigma_{xx}\) and \(\sigma_{yy}\), so it is not an independent unknown in the 2D solution).

The Governing Equations (Total: 8):

1. Equilibrium Equations (2): Identical to plane stress. \[ \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{xy}}{\partial y} = 0 \] \[ \frac{\partial \sigma_{xy}}{\partial x} + \frac{\partial \sigma_{yy}}{\partial y} = 0 \]
2. Kinematic (Strain-Displacement) Equations (3): Identical to plane stress. \[ \epsilon_{xx} = \frac{\partial u}{\partial x} \] \[ \epsilon_{yy} = \frac{\partial v}{\partial y} \] \[ \gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \]
3. Constitutive Equations (Hooke’s Law for Plane Strain) (3): These relations are different from plane stress because they incorporate the effect of the non-zero \(\sigma_{zz}\). They are often written using effective elastic constants. The stress-strain relations are: \[ \sigma_{xx} = \frac{E}{(1+\nu)(1-2\nu)} [(1-\nu)\epsilon_{xx} + \nu\epsilon_{yy}] \] \[ \sigma_{yy} = \frac{E}{(1+\nu)(1-2\nu)} [(1-\nu)\epsilon_{yy} + \nu\epsilon_{xx}] \] \[ \sigma_{xy} = G \gamma_{xy} \quad \left(\text{where } G = \frac{E}{2(1+\nu)}\right) \]

With 8 unknowns and 8 independent equations, the system is closed and solvable, providing a complete description of the 2D stress and strain state in the cross-section.

Plane Stress vs. Plane Strain

1. Plane Stress: This condition is assumed for thin bodies, like plates loaded in their plane. The key assumption is that the stress component perpendicular to the plate is zero throughout its thickness: \(\sigma_z = \tau_{xz} = \tau_{yz} = 0\). The stress strain relations are: \[ \epsilon_x = \frac{1}{E} (\sigma_x - \nu \sigma_y) \quad , \quad \epsilon_y = \frac{1}{E} (\sigma_y - \nu \sigma_x) \quad ,\]\[\gamma_{xy} = \frac{1}{G} \tau_{xy} \]
2. Plane Strain: This condition applies to very long or thick bodies where the geometry and loading do not vary along the length. It is assumed that the strain in the long direction is zero: \(\epsilon_z = \gamma_{xz} = \gamma_{yz} = 0\). This constraint means a stress \(\sigma_z\) can develop. Since \[\epsilon_z=0=\frac{1}{E}(\sigma_z-\nu \sigma_x-\nu \sigma_y)\] we get\[\sigma_z=\nu (\sigma_x+\sigma_y)\]Thus, the in plane stress strain relations become: \[ \epsilon_x = \frac{1-\nu^2}{E} \left( \sigma_x - \frac{\nu}{1-\nu} \sigma_y \right) \quad , \quad \epsilon_y = \frac{1-\nu^2}{E} \left( \sigma_y - \frac{\nu}{1-\nu} \sigma_x \right),\]\[\gamma_{xy} = \frac{1}{G} \tau_{xy} \] where \(G = \dfrac{E }{2(1+\nu)}\) is the shear modulus.

We can write both cases as \[ \begin{aligned} \epsilon_x=\frac{1}{E'}(\sigma_x-\nu'\sigma_y)&\qquad \epsilon_y=\frac{1}{E'}(\sigma_y-\nu'\sigma_x)\\[9pt] \gamma_{xy}=&\frac{2(1+\nu')}{E'}\sigma_{xy} \end{aligned} \] where \(E'\) and \(\nu'\) are effective Young’s modulus and Poisson’s ration given by the following table.