Hooke's law in 3D — generalized stress-strain (tensor): Question bank

Notation reminder so nothing here is unearned:

$σ_{ij}$ = stress (force per area; first index = face normal, second = force direction).
$ε_{ij}$ = tensorial strain (fractional deformation, symmetric $3 \times 3$ ).
Sign convention (used everywhere on this page): tension positive, compression negative. That is why a pressure $p$ pushing inward enters as $σ_{xx} = - p$ (see the hydrostatic items).
$γ_{ij}$ = engineering shear strain, defined as $γ_{ij} = 2 ε_{ij}$ for $i \neq = j$ . It measures the total change of angle between two originally perpendicular fibres, while the tensor shear $ε_{ij}$ splits that angle change symmetrically between the two directions — hence the factor of 2. Look at the figure below.
$C_{ij k l}$ = stiffness tensor (the 4-index map $σ_{ij} = C_{ij k l} ε_{k l}$ ).
$S_{ij k l}$ = compliance tensor, its inverse: $ε_{ij} = S_{ij k l} σ_{k l}$ . In Voigt matrix form it is the $6 \times 6$ matrix $[S]$ whose rows 1–3 are the normal rows and rows 4–6 are the shear rows.
$E$ = Young's modulus, $ν$ = Poisson's ratio, $G = μ$ = shear modulus, $λ$ = Lamé's first parameter, $K$ = bulk modulus.
$ε_{k k} = ε_{11} + ε_{22} + ε_{33}$ = volumetric strain (trace).

Figure — Hooke's law in 3D — generalized stress-strain (tensor)

The picture on the left shows what strain keeps: a rotation slides the block rigidly (angles unchanged, no stretch) so it stores no energy and is thrown away; a shear tilts the block (the corner angle actually changes) so it is strain. The right panel shows Poisson contraction — pull in $x$ , the block thins in $y$ .

This figure lays out the $6 \times 6$ Voigt compliance matrix $[S]$ : the top-left $3 \times 3$ block couples normal stresses to normal strains (that is where the $- ν$ Poisson terms live), while the bottom-right block is diagonal — each shear stress feeds only its own shear strain. That structural separation is exactly why "shear rows are decoupled from the normals" in the questions below.

True or false — justify

A single stiffness number like

E

is enough to describe any isotropic 3D material.

False — even isotropic solids need two constants (

E, ν

λ, μ

), because pulling in one direction also changes the other two via the Poisson effect, which

E

alone cannot encode.

The stress tensor has 9 independent components.

False — moment balance forces

σ_{ij} = σ_{j i}

, so only 6 are independent; the off-diagonal shears come in equal pairs.

A rigid-body rotation of a loaded part increases its stored elastic energy.

False — the antisymmetric part of the displacement gradient is pure rotation and is discarded when forming

ε

, so rotation contributes zero strain and zero elastic energy.

For a fully anisotropic solid,

C_{ij k l}

has 81 truly independent numbers.

False — the two minor symmetries plus the major symmetry (from energy being a scalar) cut

81 \to 36 \to 21

; only 81 positions exist, not 81 freedoms.

The major symmetry

C_{ij k l} = C_{k l ij}

holds for every linear-elastic material regardless of energy considerations.

False — it requires a conservative material where

σ_{ij} = \partial W / \partial ε_{ij}

; then mixed partials commute. A hypothetical non-conservative "linear" material could violate it.

In pure shear (

τ_{x y} = τ_{0}

only), the material changes volume.

False — the shear rows of the compliance matrix

[S]

are decoupled from the normal rows, so

ε_{k k} = 0

; shear is pure shape change, no dilation.

Hydrostatic pressure produces shear strain.

False — equal normal stresses

σ_{xx} = σ_{y y} = σ_{z z} = - p

(compression negative) give only equal normal strains and zero off-diagonal terms, so no shear, just uniform volume change.

If you pull a rod in

x

, its length in

y

genuinely gets shorter.

True — with

ν > 0

ε_{y y} = - ν σ_{0} / E < 0

, the Poisson contraction; the rod physically thins.

The two Lamé parameters

λ

and

μ

are both always positive.

False in general for

λ

—

μ = G > 0

always for stability, but

λ = \frac{E ν}{( 1 + ν ) ( 1 - 2 ν )}

becomes ==negative when

ν < 0

== (auxetic materials), which is physically allowed.

Spot the error

"Since

τ = G γ

looks just like

σ = E ε

, the compliance entry for shear must be

1/ G

with no factor of 2."

The tensor shear is

ε_{x y} = \frac{1}{2} γ_{x y}

; the compliance entry

2 (1 + ν) / E = 1/ G

only closes if you place engineering shear

γ_{x y}

(not tensor

ε_{x y}

) in the Voigt vector.

ε_{k k}

means multiply the three normal strains:

ε_{11} ε_{22} ε_{33}

No — a repeated index is a sum (Einstein convention), so

ε_{k k} = ε_{11} + ε_{22} + ε_{33}

, the trace and fractional volume change.

"In

σ_{ij} = λ ε_{k k} δ_{ij} + 2 μ ε_{ij}

, the

λ ε_{k k} δ_{ij}

term contributes to shear stress

σ_{x y}

The Kronecker delta

δ_{ij}

is zero for

i \neq = j

, so that term lives only on the diagonal; shear stresses come purely from

2 μ ε_{ij}

"To relate two

3 \times 3

tensors linearly I just need another

3 \times 3

matrix."

A matrix maps a vector to a vector; mapping one 2-index object to another linearly needs an object with 4 indices, hence

C_{ij k l}

"For a thin pressure-vessel wall, axial strain uses only the axial stress:

ε_{z} = σ_{z} / E

You dropped the Poisson pull-in from hoop stress:

ε_{z} = \frac{1}{E} [σ_{z} - ν σ_{θ}]

, which matters when sizing bolted flanges.

"Because

C_{ij k l} = C_{j ik l}

, we also automatically have

C_{ij k l} = C_{ik j l}

(swap middle indices)."

No — the minor symmetries only swap within the first pair (

ij

) or within the second pair (

k l

); swapping across pairs is not a symmetry and would be false.

"Plane strain just means the same thing as plane stress — pick whichever is easier."

They are opposite constraints: plane stress sets ==out-of-plane stress

σ_{z z} = 0

(thin plate), plane strain sets out-of-plane strain

ε_{z z} = 0

== (long body); using one where the other applies gives the wrong stiffness.

Why questions

Why do only Kronecker-delta combinations appear in the isotropic stiffness tensor?

Isotropy means

C

must look identical after any rotation; the only rotation-invariant 4th-rank tensors are built from

δ_{ij} δ_{k l}

δ_{ik} δ_{j l}

δ_{i l} δ_{j k}

Why do the last two delta terms merge into a single coefficient

μ

Minor symmetry (

ε_{k l} = ε_{l k}

) forces

δ_{ik} δ_{j l}

and

δ_{i l} δ_{j k}

to always act together, so they share one constant.

Why do we symmetrize the displacement gradient when defining strain?

The antisymmetric part is rotation (no stretch, stores no energy); only the symmetric part changes lengths, so strain keeps just that.

Why does energy conservation give the major symmetry?

Stress is the gradient of stored energy

σ_{ij} = \partial W / \partial ε_{ij}

, so

C_{ij k l} = \partial^{2} W / \partial ε_{ij} \partial ε_{k l}

; mixed partials commute, forcing

C_{ij k l} = C_{k l ij}

Why can't

ν

exceed

\frac{1}{2}

for a stable ordinary material?

Then

K = \frac{E}{3 ( 1 - 2 ν )}

would be negative, meaning the material expands under compression — a self-amplifying instability.

Why does the shear modulus

G

, not

E

, govern the twist of a spacecraft drive shaft?

Twisting is a pure shear deformation; the shear compliance depends on

G = E / [2 (1 + ν)]

, independent of the normal-stress response.

Why do composite panels need up to 21 constants while aluminium needs 2?

Composites have direction-dependent stiffness (no rotational symmetry), so their

C_{ij k l}

keeps all 21 anisotropic constants; isotropy collapses this to 2.

Why is the distinction between

ε_{ij}

and

γ_{ij}

worth policing so carefully?

Because

γ_{ij} = 2 ε_{ij}

; silently mixing them doubles or halves your shear stress, corrupting any von Mises check that combines normal and shear terms.

Edge cases

What happens to

K

ν \to \frac{1}{2}

K = \frac{E}{3 ( 1 - 2 ν )} \to \infty

: the material becomes incompressible (like rubber), resisting any volume change while still shearing freely.

What does

ν = 0

physically mean?

Pulling in

x

produces no lateral contraction (

ε_{y y} = ε_{z z} = 0

); cork approximates this, useful when you don't want a stretched part to pinch its neighbours.

Is a negative

ν

physically allowed, and what is such a material called?

Yes — down to

ν > - 1

for stability; these are auxetic materials that get fatter sideways when stretched (some foams and lattices).

What does the uniaxial load (

σ_{xx} = σ_{0}

, rest zero) reduce the 3D law to?

The cross terms vanish, recovering 1D Hooke

ε_{xx} = σ_{0} / E

, while still predicting the lateral

ε_{y y} = ε_{z z} = - ν σ_{0} / E

For plane stress in a thin vessel wall, which stress is set to zero and why?

The ==out-of-plane (through-thickness) stress

σ_{z z} \approx 0

==, because a thin wall has nothing to push back against across its small thickness.

In plane strain, which quantity is zero and when does it apply?

The ==out-of-plane strain

ε_{z z} = 0

==, applying to long bodies (dam, buried pipe, long weld) whose ends can't move axially; note this forces a non-zero constraint stress

σ_{z z} = ν (σ_{xx} + σ_{y y})

Why does plane strain generate an out-of-plane stress even though the strain there is zero?

With

ε_{z z} = 0

the material wants to Poisson-contract in

z

but is prevented, so the constraint pushes back as ==

σ_{z z} = ν (σ_{xx} + σ_{y y})

==.

At the incompressible limit, what happens to Lamé's

λ

λ = \frac{E ν}{( 1 + ν ) ( 1 - 2 ν )} \to \infty

ν \to \frac{1}{2}

, which is why standard displacement finite elements "lock" near incompressibility and need special formulations.

Recall One-line self-test

Cover every answer above and re-run the "Spot the error" section only — those are the traps most likely to bite in an exam or a design review.