Strain Tensor

What does ε_xx measure? Take a line element dx in the x-direction. After deformation, its length changes by d(u_x) = (∂u_x/∂x)dx. The relative length change is ΔL/L = ∂u_x/∂x = ε_xx.

ΔV/V = tr(E): For small strains, the volume change of a cube with edges (1+ε_xx)(1+ε_yy)(1+ε_zz) ≈ 1 + ε_xx + ε_yy + ε_zz. The trace of the strain tensor is the volumetric strain!

Poisson effect: In real materials, stretching in one direction causes contraction in others: ε_yy = ε_zz = −νε_xx, where ν is Poisson's ratio (ν ≈ 0.3 for steel, ν = 0.5 for rubber = incompressible).

Pure shear vs simple shear:
Pure shear: ε_xy ≠ 0, all other components zero. The body deforms symmetrically — the cube becomes a rhombus (Raute). Both x- and y-edges tilt by γ/2 toward each other.

Simple shear (einfache Scherung): Only u_x = γ · y, i.e., layers slide over each other. This gives ∇u with only one nonzero off-diagonal entry. But E = ½(∇u + (∇u)^T) still symmetrizes it, so simple shear = pure shear + rotation!

Key insight: The strain tensor cannot distinguish between simple shear and pure shear — the antisymmetric part (rotation) is "factored out."

What does Mohr's circle show?
Imagine cutting through the material at angle θ. On that cut face you measure:
• ε_n = normal strain perpendicular to the cut (stretch/compress)
• ε_s = shear strain along the cut (sliding)

Rotate θ from 0° to 180° — the points (ε_n, ε_s) trace Mohr's circle. This gives you ALL possible strain states for every cut direction in one picture.

Key points on the circle:
• Principal strains = where the circle crosses the ε_n axis (ε_s=0 → no shear!)
• Principal directions = eigenvectors = cut angles where shear vanishes
• Max shear = top/bottom of circle = at 45° to principal axes
• Starting points: A=(ε_xx, ε_xy) for θ=0 and B=(ε_yy, −ε_xy) for θ=90° — diametrically opposite!
• 2θ on the circle = double the physical angle. The point moves twice as fast on the circle as your cut rotates in space.
• Convention: ε_s positive downward on the Mohr diagram (standard mechanics sign convention).

Eigenvalues & Eigenvectors:
The eigenvalue equation det(E − λI) = 0 gives the principal strains ε₁ ≥ ε₂ ≥ ε₃. These are the strains on faces where ALL shear vanishes.

Physical meaning of principal directions:
The eigenvectors define a rotated coordinate system in which the tensor becomes diagonal:
E' = diag(ε₁, ε₂, ε₃) In these directions there is ONLY stretch/compression, no sliding! All off-diagonal entries are zero.

When can you reduce 3D to 2D?
A 3D tensor [ε]_xyz is reducible to 2D when:

Case 1: Plane strain state
ε_xz = ε_yz = 0 → the z-axis is already a principal direction!
Solve only the 2×2 xy-problem. ε₃ = ε_zz.

Case 2: Already diagonal
All ε_ij=0 (i≠j) → the coordinate axes ARE the principal axes.
No eigenvalue problem needed! ε_1,2,3 = ε_xx, ε_yy, ε_zz (sorted).

Exam tip: Always check the tensor structure first!
If a row/column has only its diagonal entry → that axis is a principal direction → 2D suffices.

Recipe: Drawing Mohr's circle
1. Compute center: C = (ε_xx+ε_yy)/2
2. Compute radius: R = √((ε_xx−ε_yy)²/4 + ε_xy²)
3. Draw circle with center (C, 0) and radius R
4. Plot starting point A=(ε_xx, ε_xy) — must lie on the circle!
5. Diametrically opposite: B=(ε_yy, −ε_xy)
6. Intersections with ε_n axis = principal strains
7. Top/bottom of circle = max shear strain
8. 2α from A to the right intersection = double the principal angle

Rotation invariants: I₁=tr(E), I₂, I₃=det(E) do NOT change under rotation — the circle just shifts on the ε_n axis.

Why decompose? In material science and plasticity, volume change and shape change are governed by different physical mechanisms:
• Volume change is resisted by the bulk modulus K = E/(3(1−2ν)). For incompressible materials (ν = 0.5), K → ∞ and no volume change is possible.
• Shape change is resisted by the shear modulus G = E/(2(1+ν)). Yielding (plastic deformation) depends only on the deviatoric part!

Von Mises criterion: Yielding occurs when the deviatoric strain energy reaches a critical value. This is why we need the decomposition — it separates the "harmful" shape change from the "harmless" volume change.

Derivation of E' = R · E · R^T:
The strain tensor maps a direction n to the strain on that face: ε_n = n^T · ε · n.

In the rotated system, the direction n' corresponds to n = R^Tn' in the original system. So:
ε_n = n^Tεn = (R^Tn')^T ε (R^Tn') = n'^T (R E R^T) n' = n'^T ε' n' Therefore E' = R E R^T. This is the similarity transformation for symmetric tensors.

Key insight: The transformation preserves eigenvalues (principal strains) and the trace (I₁). Only the components change, not the physics.

2D explicit formulas (rotation by θ about z-axis):
R = [[cosθ, sinθ], [−sinθ, cosθ]] ε'_xx = cos²θ · ε_xx + sin²θ · ε_yy + 2sinθcosθ · ε_xy ε'_yy = sin²θ · ε_xx + cos²θ · ε_yy − 2sinθcosθ · ε_xy ε'_xy = (ε_yy−ε_xx)sinθcosθ + ε_xy(cos²θ−sin²θ) These are exactly the formulas that Mohr's circle encodes graphically!

Why cylindrical coordinates? — Pressure vessels
For a thin-walled cylindrical pressure vessel (pipe, tank, boiler), the geometry has axial symmetry. Cartesian coordinates (x,y,z) would give complicated, angle-dependent components. In cylindrical coordinates (r, θ, z), the strain tensor becomes diagonal by symmetry:
[ε]_rθz = diag(ε_rr, ε_θθ, ε_zz) ε_rr = radial strain (wall thickening/thinning)
ε_θθ = hoop (circumferential) strain — the dominant component
ε_zz = axial strain (lengthening)

The transformation from Cartesian to cylindrical:
R(θ) = [[cosθ, sinθ, 0], [−sinθ, cosθ, 0], [0, 0, 1]] This is just a rotation about the z-axis by angle θ — the same R we use for 2D!

Thin-walled pressure vessel formulas:
For internal pressure p, mean radius R, wall thickness t:
σ_θθ = pR/t   (hoop stress — largest!) σ_zz = pR/(2t)   (axial stress = half of hoop) σ_rr ≈ 0   (thin wall → negligible) Using Hooke's law (E = (1/E_Y)[T − ν(tr(T)I − T)]):
ε_θθ = (pR)/(Et)(1 − ν/2) ε_zz = (pR)/(2Et)(1 − 2ν) ε_rr = −(pR)/(Et)(ν − ν/2) = −νpR/(Et)(3/2 − 1/(2)) Key point: hoop strain is always the largest — this is why vessels fail along their length (circumferential crack), not around their circumference!

Why this matters for choosing coordinates:
In Cartesian (x,y,z), the stress/strain at a point on the vessel wall depends on the angular position θ — you get messy, varying components. In cylindrical (r,θ,z), every point on the wall has the same components by symmetry. The tensor is automatically diagonal!

General rule: Always choose the coordinate system that matches the symmetry of your problem:
• Cylindrical vessel → cylindrical (r,θ,z)
• Spherical vessel → spherical (r,θ,φ)
• Rectangular beam → Cartesian (x,y,z)

In the right system, the tensor often becomes diagonal or has fewer non-zero components — simpler eigenvalue problems, cleaner formulas.