Plane stress, plane strain and compatibility equations

Why it matters

Full three-dimensional elasticity is rarely solved by hand. Most aircraft skins, webs and ribs are thin sheets, and some parts such as long thick lugs or constrained sections behave as if they cannot strain along one axis. Recognising plane stress or plane strain reduces the problem to two dimensions. Compatibility is the extra condition that makes a strain field physically possible, and it underlies the Airy stress function used for classical solutions around holes and cut-outs.

Key ideas

Plane stress. The stresses on planes parallel to one surface are zero: σz = τxz = τyz = 0. This is a good model for thin plates loaded in their own plane (wing skins, spar webs, fuselage panels). The out-of-plane strain is not zero: the plate thins or thickens by εz = −ν(σx + σy)/E.

Plane strain. The strains in one direction are zero: εz = γxz = γyz = 0. This models long bodies whose cross-section and loading do not vary along z and whose ends are restrained (long dams, tunnels, long pipes with restrained ends, and the interior of very thick sections near a crack tip). The out-of-plane stress is not zero: σz = ν(σx + σy) is needed to stop the Poisson strain.

Both are two-dimensional problems with three in-plane unknown stresses σx, σy, τxy. They differ only in the stress–strain law. Plane-strain formulas can be obtained from plane-stress ones by replacing E with E/(1 − ν²) and ν with ν/(1 − ν). In plane strain the material is effectively stiffer.

Strain–displacement relations. With displacements u (along x) and v (along y): εx = ∂u/∂x, εy = ∂v/∂y and γxy = ∂u/∂y + ∂v/∂x. These are valid for small strains.

Compatibility. Three strain components come from only two displacement functions, so they cannot be chosen freely. Eliminating u and v gives the 2-D compatibility equation ∂²εx/∂y² + ∂²εy/∂x² = ∂²γxy/∂x∂y. If it fails, no continuous single-valued displacement field exists: the body would have to crack open or overlap. Any strains computed from a real displacement field satisfy it automatically. In 3-D there are six such equations.

The full 2-D elasticity problem combines three sets of equations: equilibrium (∂σx/∂x + ∂τxy/∂y = 0 and ∂τxy/∂x + ∂σy/∂y = 0, with no body force), compatibility, and the stress–strain law, together with boundary conditions.

Airy stress function. Choose φ(x, y) with σx = ∂²φ/∂y², σy = ∂²φ/∂x² and τxy = −∂²φ/∂x∂y. Equilibrium is then satisfied automatically, and compatibility reduces to the biharmonic equation ∇⁴φ = 0. Any polynomial of degree three or less satisfies it; higher polynomials must be checked. Classical results for beams, plates with holes and stress concentration factors come from this approach.

Formulas

εx = (σx − ν·σy)/E, εy = (σy − ν·σx)/E, εz = −ν·(σx + σy)/E, γxy = τxy/G Plane stress (σz = 0). Strains dimensionless, stresses and E, G in Pa.

σz = ν·(σx + σy) Plane strain (εz = 0).

εx = (1 + ν)·[(1 − ν)·σx − ν·σy]/E, εy = (1 + ν)·[(1 − ν)·σy − ν·σx]/E Plane strain in-plane strains.

εx = ∂u/∂x, εy = ∂v/∂y, γxy = ∂u/∂y + ∂v/∂x Small-strain kinematics; u, v in m.

∂²εx/∂y² + ∂²εy/∂x² = ∂²γxy/(∂x·∂y) 2-D compatibility equation.

σx = ∂²φ/∂y², σy = ∂²φ/∂x², τxy = −∂²φ/(∂x·∂y) Airy stress function φ (units Pa·m², so that its second derivatives are stresses in Pa).

∂⁴φ/∂x⁴ + 2·∂⁴φ/(∂x²·∂y²) + ∂⁴φ/∂y⁴ = 0 Biharmonic equation (no body force).

Worked examples

Example 1 (standard). A steel element (E = 200 GPa, ν = 0.3) carries σx = 100 MPa and σy = 40 MPa with no in-plane shear. Find all normal strains and σz (a) in plane stress and (b) in plane strain.

1. Plane stress: εx = (σx − νσy)/E = (100 − 12)/200 000 = 4.40 × 10⁻⁴.
2. εy = (σy − νσx)/E = (40 − 30)/200 000 = 5.0 × 10⁻⁵.
3. εz = −ν(σx + σy)/E = −0.3 × 140/200 000 = −2.10 × 10⁻⁴; σz = 0.
4. Plane strain: σz = ν(σx + σy) = 0.3 × 140 = 42 MPa; εz = 0.
5. εx = [σx − ν(σy + σz)]/E = (100 − 0.3 × 82)/200 000 = 75.4/200 000 = 3.77 × 10⁻⁴.
6. εy = [σy − ν(σx + σz)]/E = (40 − 0.3 × 142)/200 000 = −2.6/200 000 = −1.3 × 10⁻⁵.
7. Check with the plane-strain formula: (1.3/200 000) × (0.7 × 100 − 0.3 × 40) = 6.5 × 10⁻⁶ × 58 = 3.77 × 10⁻⁴.

Answer: plane stress: εx = 4.40 × 10⁻⁴, εy = 5.0 × 10⁻⁵, εz = −2.10 × 10⁻⁴; plane strain: σz = 42 MPa, εx = 3.77 × 10⁻⁴, εy = −1.3 × 10⁻⁵.

Example 2 (GATE level). (a) A proposed strain field is εx = k·y², εy = k·x², γxy = c·x·y with k constant. For what c is it compatible? (b) Show that φ = A·(x³·y − 3·x·y³) is a valid Airy stress function and find the stresses at (x, y) = (1 m, 0.5 m) for A = 2 MPa/m².

1. (a) ∂²εx/∂y² = 2k and ∂²εy/∂x² = 2k, so the left side is 4k.
2. ∂²γxy/∂x∂y = c. Compatibility requires c = 4k.
3. (b) ∂⁴φ/∂x⁴ = 0 and ∂⁴φ/∂y⁴ = 0 (φ is only cubic in each variable). φ_xx = 6A·x·y, so ∂⁴φ/∂x²∂y² = 0. Hence ∇⁴φ = 0: valid.
4. σx = ∂²φ/∂y² = −18A·x·y = −18 × 2 × 1 × 0.5 = −18 MPa.
5. σy = ∂²φ/∂x² = 6A·x·y = 6 × 2 × 0.5 = 6 MPa.
6. τxy = −∂²φ/∂x∂y = −(3A·x² − 9A·y²) = −(6 − 4.5) = −1.5 MPa.
7. Check equilibrium in general: ∂σx/∂x + ∂τxy/∂y = −18A·y + 18A·y = 0, and ∂τxy/∂x + ∂σy/∂y = −6A·x + 6A·x = 0.

Answer: (a) c = 4k; (b) σx = −18 MPa, σy = 6 MPa, τxy = −1.5 MPa at (1, 0.5).

Common mistakes

• Assuming εz = 0 in plane stress, or σz = 0 in plane strain. Each condition zeroes one quantity, not both.
• Calling a thin sheet "plane strain"; it is the classic plane-stress case.
• Writing the compatibility equation with 2∂²γxy/∂x∂y; that factor belongs to the tensor shear strain form.
• Forgetting the minus sign in τxy = −∂²φ/∂x∂y.
• Accepting a polynomial Airy function of degree four or more without checking ∇⁴φ = 0.

For GATE AE

Expect conceptual questions distinguishing plane stress and plane strain, quick numericals on εz in plane stress or σz in plane strain, checks of whether a given strain field is compatible or a given Airy function is admissible, and stresses from a given Airy function. Practise partial differentiation of polynomials quickly and always verify equilibrium as a check.

Quick check

1. In plane stress, which stress components are zero?
2. In plane strain with σx = 50 MPa, σy = 30 MPa and ν = 0.25, what is σz?
3. Is εx = 3x, εy = 2y, γxy = 0 compatible?
4. Is φ = x²y² a valid Airy function?

Answers: 1. σz, τxz and τyz. 2. 20 MPa. 3. Yes (all second derivatives in the equation are zero). 4. No: ∇⁴φ = 2 × 4 = 8 ≠ 0.