平面応力問題
Theory and Physics
What is Plane Stress?
Professor, does "plane stress" mean reducing a 3D problem to 2D?
Yes. Solving a 3D elastic body results in a vast number of degrees of freedom. However, for thin plate-like structures subjected only to in-plane loads, the stress components in the thickness direction can be considered zero. This is the plane stress assumption.
What kind of structures fit this assumption?
Structures where the thickness is sufficiently thin compared to the in-plane dimensions. Specifically:
- Thin plates subjected to in-plane loads (tension, compression, shear)
- Analysis of brackets and sheet metal parts
- Thin walls of dams under water pressure
Conversely, when can plane stress not be used?
When there is constraint in the thickness direction. For example, thick walls of dams or cross-sections of long tunnels. In such cases, the plane strain assumption (strain in the thickness direction is zero) is used. Plane stress and plane strain are similar but distinct; confusing them leads to large errors.
Governing Equations
Please tell me the governing equations for plane stress.
2D force equilibrium equations:
Compatibility condition (strain compatibility):
What about the constitutive law (Hooke's law)?
Hooke's law for plane stress (matrix form):
$E/(1-\nu^2)$ is larger than the usual $E$, right? Because lateral deformation is constrained by the Poisson effect, the effective stiffness increases.
Good point. However, in plane stress, deformation in the lateral direction ($z$ direction) is free. $E/(1-\nu^2)$ appears because the two stress components are coupled, which has a slightly different meaning from the 3D $E$.
Important point: In plane stress, $\sigma_{zz} = 0$, but $\varepsilon_{zz} \neq 0$:
Strain occurs in the thickness direction. When the plate is stretched in-plane, it becomes thinner — this is the Poisson effect.
Plane Stress vs. Plane Strain
Please explain the difference from plane strain in more detail.
Characteristic Plane Stress Plane Strain Assumption $\sigma_{zz} = 0$ $\varepsilon_{zz} = 0$ Applicable Object Thin plates (thickness << in-plane dimensions) Long structures (depth >> cross-sectional dimensions) Deformation in $z$ direction Exists ($\varepsilon_{zz} \neq 0$) None Stress in $z$ direction None Exists ($\sigma_{zz} = \nu(\sigma_x + \sigma_y)$) Effective Stiffness $E/(1-\nu^2)$ $E(1-\nu)/((1+\nu)(1-2\nu))$ Typical Example Sheet metal bracket Dam cross-section, long tunnel
The effective stiffness of plane strain is larger than that of plane stress... meaning deformation is smaller under the same load.
Yes. Because deformation in the $z$ direction is completely constrained, the stiffness increases due to the 3D constraint effect. This difference is about 10% for $\nu = 0.3$. It may seem small, but it's a non-negligible difference in stress evaluation.
Airy Stress Function
I heard there is a solution method using a "stress function" for plane problems.
Using the Airy stress function $\phi(x,y)$, the equilibrium equations and compatibility condition can be combined into a single equation:
$$ \nabla^4 \phi = 0 \quad \text{(without gravity)} $$
Please explain the difference from plane strain in more detail.
| Characteristic | Plane Stress | Plane Strain |
|---|---|---|
| Assumption | $\sigma_{zz} = 0$ | $\varepsilon_{zz} = 0$ |
| Applicable Object | Thin plates (thickness << in-plane dimensions) | Long structures (depth >> cross-sectional dimensions) |
| Deformation in $z$ direction | Exists ($\varepsilon_{zz} \neq 0$) | None |
| Stress in $z$ direction | None | Exists ($\sigma_{zz} = \nu(\sigma_x + \sigma_y)$) |
| Effective Stiffness | $E/(1-\nu^2)$ | $E(1-\nu)/((1+\nu)(1-2\nu))$ |
| Typical Example | Sheet metal bracket | Dam cross-section, long tunnel |
The effective stiffness of plane strain is larger than that of plane stress... meaning deformation is smaller under the same load.
Yes. Because deformation in the $z$ direction is completely constrained, the stiffness increases due to the 3D constraint effect. This difference is about 10% for $\nu = 0.3$. It may seem small, but it's a non-negligible difference in stress evaluation.
I heard there is a solution method using a "stress function" for plane problems.
Using the Airy stress function $\phi(x,y)$, the equilibrium equations and compatibility condition can be combined into a single equation:
Solving this biharmonic equation, the stress components are:
So the stresses automatically satisfy equilibrium. Convenient!
Classical elasticity problems (infinite plate with a hole, wedge, half-plane contact problem) can be solved with the Airy function. The Kirsch problem (stress concentration around a circular hole in an infinite plate) is also a solution using the Airy function. It's very useful as a verification problem for FEM.
Summary
Let me organize the theory of plane stress.
Key points:
- Assumption of $\sigma_{zz} = 0$ — Applied to in-plane problems of thin plates
- Distinction from plane strain is crucial — Using the wrong assumption leads to errors of 10% or more
- Constitutive law includes $E/(1-\nu^2)$ — 2D coupling effect
- $\varepsilon_{zz} \neq 0$ — Strain exists in the thickness direction
- Classical problems can be solved with the Airy function — Useful for FEM verification
To "correctly" reduce a 3D problem to 2D, understanding the physics is essential, isn't it?
Exactly. 2D elements can dramatically reduce computational cost, but the prerequisite is to correctly judge when the assumption holds. If in doubt, solve it in 3D and compare with the 2D results.
Theoretical Basis of the Plane Stress Assumption
The plane stress assumption (σz=τyz=τxz=0) is applied to thin plates where "the thickness is sufficiently small compared to the in-plane dimensions." Kirchhoff organized the framework for separating bending and extension problems of plates in 1909, which became the analytical foundation for subsequent thin plate and membrane problems. For parts like aircraft wing spar webs and automotive door panels with thicknesses around 1-3mm, 2D plane stress models often agree with 3D shell analyses within an error margin of 1%.
Physical Meaning of Each Term
- Inertia term (mass term): $\rho \ddot{u}$, i.e., "mass × acceleration". Have you ever experienced being thrown forward when slamming on the brakes? That "feeling of being carried away" is precisely the inertial force. Heavier objects are harder to set in motion and harder to stop once moving. Buildings shake during earthquakes because the ground moves suddenly while the building's mass "gets left behind." In static analysis, this term is set to zero, which assumes "forces are applied slowly so acceleration can be ignored." It absolutely cannot be omitted for impact loads or vibration problems.
- Stiffness term (elastic restoring force): $Ku$ or $\nabla \cdot \sigma$. When you pull a spring, you feel a "force trying to return it," right? That is Hooke's law $F=kx$, the essence of the stiffness term. Now a question — an iron rod and a rubber band, which stretches more under the same force? Obviously the rubber. This "resistance to stretching" is the Young's modulus $E$, which determines stiffness. A common misconception: "High stiffness ≠ strong." Stiffness is "resistance to deformation," strength is "resistance to failure" — they are different concepts.
- External force term (load term): Body force $f_b$ (e.g., gravity) and surface force $f_s$ (pressure, contact force). Think of it this way — the weight of a truck on a bridge is a "force acting on the entire volume" (body force), while the force of the tires pushing on the road surface is a "force acting only on the surface" (surface force). Wind pressure, water pressure, bolt tightening force... all are external forces. A common mistake here: getting the load direction wrong. Intending "tension" but it becomes "compression" — sounds like a joke, but it actually happens when coordinate systems are rotated in 3D space.
- Damping term: Rayleigh damping $C\dot{u} = (\alpha M + \beta K)\dot{u}$. Try plucking a guitar string. Does the sound continue forever? No, it gradually fades. That's because vibrational energy is converted to heat by air resistance and internal friction in the string. Car shock absorbers work on the same principle — they intentionally absorb vibrational energy to improve ride comfort. What if damping were zero? Buildings would continue shaking forever after an earthquake. Since that doesn't happen in reality, setting appropriate damping is important.
Assumption Conditions and Applicability Limits
- Continuum assumption: Treats material as a continuous medium, ignoring microscopic heterogeneity
- Small deformation assumption (for linear analysis): Deformation is sufficiently small compared to initial dimensions, and the stress-strain relationship is linear
- Isotropic material (unless otherwise specified): Material properties are independent of direction (anisotropic materials require separate tensor definitions)
- Quasi-static assumption (for static analysis): Ignores inertial and damping forces, considering only the balance between external and internal forces
- Non-applicable cases: For large deformation/large rotation problems, geometric nonlinearity is required. For nonlinear material behavior like plasticity and creep, constitutive law extensions are needed.
Dimensional Analysis and Unit Systems
| Variable | SI Unit | Notes / Conversion Memo |
|---|---|---|
| Displacement $u$ | m (meter) | When inputting in mm, unify load and elastic modulus to MPa/N system |
| Stress $\sigma$ | Pa (Pascal) = N/m² | MPa = 10⁶ Pa. Be careful of unit system inconsistency when comparing with yield stress |
| Strain $\varepsilon$ | Dimensionless (m/m) | Note the distinction between engineering strain and logarithmic strain (for large deformations) |
| Elastic Modulus $E$ | Pa | Steel: ~210 GPa, Aluminum: ~70 GPa. Note temperature dependence |
| Density $\rho$ | kg/m³ | In mm system: tonne/mm³ (= 10⁻⁹ tonne/mm³ for steel) |
| Force $F$ | N (Newton) | Unify to N in mm system, N in m system |
Numerical Solution Methods and Implementation
Plane Stress Analysis by FEM
When solving plane stress problems with FEM, what kind of elements are used?
2D plane stress elements. Each node has 2 degrees of freedom ($u_x, u_y$), requiring about 1/3 the DOF of 3D solid elements.
Typical Element Types
| Element | Number of Nodes | Shape Function | Accuracy | Application |
|---|---|---|---|---|
| 3-node triangle (CST) | 3 | Linear | Low (constant strain) | Filling for automatic meshing |
| 6-node triangle (LST) | 6 | Quadratic | High | Automatic meshing for complex shapes |
| 4-node quadrilateral (Q4) | 4 | Bilinear | Medium | Regular mesh |
| 8-node quadrilateral (Q8) | 8 | Quadratic | High | Standard for precise analysis |
CST (Constant Strain Triangle) has constant strain within the element, as the name suggests... so it can't represent stress gradients, right?
Exactly. CST has significantly low accuracy in areas with large stress gradients like stress concentration zones. It can only be used for educational purposes or rough screening. In practice, quadratic elements (6-node triangle or 8-node quadrilateral) are standard.
Influence of Integration Scheme
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