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

WHY do we need a tensor at all?

WHAT is being related? Two quantities that each need direction information:

Stress $\sigma_{ij}$ : force per area. First index = face normal, second index = force direction. 9 components (6 independent, since $\sigma_{ij}=\sigma_{ji}$ by moment balance).
Strain $\varepsilon_{ij}$ : fractional deformation. Also symmetric, 6 independent.

WHY not just a matrix? Because at each point the state of loading is itself a $3\times3$ object. Relating one $3\times3$ tensor linearly to another requires an object with 4 indices:

\boxed{\ \sigma_{ij} = C_{ijkl}\,\varepsilon_{kl}\ }\qquad (i,j,k,l\in\{1,2,3\})

Summation over repeated $k,l$ (Einstein convention). This is Hooke's law in its most general linear-elastic form.

Deriving strain from displacement (first principles)

Why this form? Take two nearby points separated by $d\mathbf{x}$ . After deformation the separation becomes $dx_i + \frac{\partial u_i}{\partial x_j}dx_j$ . The change in squared length, to first order, involves the symmetric part of $\partial u_i/\partial x_j$ . The antisymmetric part is pure rotation (no stretch), so we discard it — that's why we symmetrize. Rotation doesn't store elastic energy, so it must not appear in $\varepsilon$ .

Counting the constants: 81 → 21 → 2

Intuition Why so many indices collapse

$C_{ijkl}$ has $3^4=81$ components. But symmetries slash this:

$\sigma_{ij}=\sigma_{ji}\Rightarrow C_{ijkl}=C_{jikl}$ (minor sym)
$\varepsilon_{kl}=\varepsilon_{lk}\Rightarrow C_{ijkl}=C_{ijlk}$ (minor sym)
Elastic energy $W=\tfrac12 C_{ijkl}\varepsilon_{ij}\varepsilon_{kl}$ is a scalar, so $C_{ijkl}=C_{klij}$ (major sym, because $\partial^2 W/\partial\varepsilon_{ij}\partial\varepsilon_{kl}$ is symmetric in the pair swap).

Result: $81\to 36 \to \boxed{21}$ independent constants for a fully anisotropic solid. Add isotropy (same in every direction) $\Rightarrow$ only $\boxed{2}$ constants survive.

WHY does energy give the major symmetry? For a conservative elastic material stress is the gradient of a stored-energy density: $\sigma_{ij}=\partial W/\partial\varepsilon_{ij}$ . Then $C_{ijkl}=\partial^2 W/\partial\varepsilon_{ij}\partial\varepsilon_{kl}$ , and mixed partials commute — hence swapping the $ij$ and $kl$ pairs leaves $C$ unchanged.

The isotropic law (the workhorse)

Here $\varepsilon_{kk}=\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}$ is the volumetric strain (trace = fractional volume change). $\lambda$ and $\mu$ are the Lamé parameters; $\mu\equiv G$ is the shear modulus.

Why only two building blocks? Isotropy means $C$ must look the same after any rotation — an isotropic tensor. The only isotropic 4th-rank tensors are combinations of $\delta_{ij}\delta_{kl}$ , $\delta_{ik}\delta_{jl}$ , $\delta_{il}\delta_{jk}$ . Minor symmetry forces the last two to appear together, leaving two coefficients.

Inverting to strain-from-stress (engineering form)

Solving the above for $\varepsilon$ gives the form students memorize:

\varepsilon_{ij} = \frac{1+\nu}{E}\sigma_{ij} - \frac{\nu}{E}\sigma_{kk}\,\delta_{ij}

Component-wise (normal strains):

\varepsilon_{xx} = \frac1E\big[\sigma_{xx}-\nu(\sigma_{yy}+\sigma_{zz})\big]

The $-\nu(\ldots)$ term is the Poisson effect: stress in $y,z$ shrinks $x$ .

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

Voigt (matrix) notation — how engineers actually compute

Because both tensors are symmetric, pack their 6 independent components into vectors:

\{\sigma\}=[\sigma_{xx},\sigma_{yy},\sigma_{zz},\tau_{yz},\tau_{xz},\tau_{xy}]^T,\quad \{\varepsilon\}=[\varepsilon_{xx},\varepsilon_{yy},\varepsilon_{zz},\gamma_{yz},\gamma_{xz},\gamma_{xy}]^T

with engineering shear $\gamma_{xy}=2\varepsilon_{xy}$ . Then $\{\sigma\}=[C]\{\varepsilon\}$ with a $6\times6$ matrix. The isotropic compliance $[S]=[C]^{-1}$ :

[S]=\frac1E\begin{bmatrix} 1 & -\nu & -\nu & 0&0&0\\ -\nu & 1 & -\nu & 0&0&0\\ -\nu & -\nu & 1 & 0&0&0\\ 0&0&0& 2(1+\nu)&0&0\\ 0&0&0&0&2(1+\nu)&0\\ 0&0&0&0&0&2(1+\nu) \end{bmatrix}

Worked examples

Worked example 2 — Hydrostatic pressure (find

$K$ from scratch) Load: $\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=-p$ , no shear. $\varepsilon_{xx}=\tfrac1E[-p-\nu(-p-p)]=\tfrac{-p(1-2\nu)}{E}.$ Why? Each normal strain feels its own $-p$ minus Poisson pull-in from the other two. Volume change $\varepsilon_{kk}=3\varepsilon_{xx}=\dfrac{-3p(1-2\nu)}{E}$ . Define $K=-p/\varepsilon_{kk}=\dfrac{E}{3(1-2\nu)}$ . ✔ Insight: if $\nu\to\tfrac12$ , $K\to\infty$ — incompressible (rubber). $\nu>\tfrac12$ would give negative $K$ (unstable), so $\nu<\tfrac12$ always.

Recall Feynman: explain to a 12-year-old

Imagine a foam block. If you squeeze it from the top, it doesn't just get shorter — it also bulges out the sides. So "how much it squishes" depends on which sides you push and how hard, all at the same time. One number can't describe that. So we use a big table of numbers (a tensor) that says: "if you push this way, here's exactly how it changes in every direction." For most simple materials the whole giant table boils down to just two numbers: how stiff it is ( $E$ ) and how much it bulges sideways ( $\nu$ ). That's the whole trick.

Flashcards

General 3D Hooke's law in index form

\sigma_{ij}=C_{ijkl}\varepsilon_{kl}

, sum over

k,l

How many independent components in a fully anisotropic stiffness tensor, and why?

21; from

3^4=81

reduced by two minor symmetries (

\sigma,\varepsilon

symmetric) to 36, then major symmetry (energy) to 21.

What reduces the 21 constants to 2?

Isotropy — invariance under all rotations, leaving only Lamé

\lambda

and

\mu

Isotropic stress–strain law (Lamé form)

\sigma_{ij}=\lambda\,\varepsilon_{kk}\delta_{ij}+2\mu\,\varepsilon_{ij}

Definition of small-strain tensor

\varepsilon_{ij}=\tfrac12(\partial_j u_i+\partial_i u_j)

; symmetric part of displacement gradient (rotation removed).

Why do we symmetrize the displacement gradient?

The antisymmetric part is rigid rotation, which stores no elastic energy.

Where does major symmetry

C_{ijkl}=C_{klij}

come from?

\sigma_{ij}=\partial W/\partial\varepsilon_{ij}

, so

C=\partial^2 W/\partial\varepsilon\partial\varepsilon

and mixed partials commute.

Engineering strain-from-stress for normal component

\varepsilon_{xx}=\tfrac1E[\sigma_{xx}-\nu(\sigma_{yy}+\sigma_{zz})]

Shear modulus in terms of

E,\nu

G=\mu=E/[2(1+\nu)]

Bulk modulus and physical limit

K=E/[3(1-2\nu)]

;

\nu<1/2

for positive

K

;

\nu\to1/2

= incompressible.

The factor-of-2 shear trap

Engineering shear

\gamma_{xy}=2\varepsilon_{xy}

; compliance uses

\gamma

so shear entry

=1/G=2(1+\nu)/E

Connections

Stress tensor and Cauchy's relation
Strain tensor and displacement gradient
Elastic strain energy density
Poisson's ratio and material limits
Isotropic vs anisotropic materials (composites)
Thin-walled pressure vessels (spacecraft tanks)
Von Mises yield criterion
Finite Element Method — stiffness matrices

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, 1D mein Hooke's law simple hai: $\sigma = E\varepsilon$ , matlab jitna kheencho utna khinchta hai. Lekin real material — jaise spacecraft ka aluminium panel — ko aap ek saath multiple directions mein push aur shear kar sakte ho. Aur ek important baat: jab aap $x$ direction mein kheencho, to material $y$ aur $z$ mein patla ho jaata hai — isko Poisson effect kehte hain. Isliye ek single number $E$ kaafi nahi, humein ek tensor chahiye jo poora relationship bataaye: $\sigma_{ij} = C_{ijkl}\varepsilon_{kl}$ .

Ab $C_{ijkl}$ mein $3^4 = 81$ components hote hain, lekin symmetries ki wajah se ye ghat ke 21 reh jaate hain (anisotropic material ke liye). Aur agar material isotropic ho — har direction mein same behaviour — to sirf do constants bachte hain: Lamé ke $\lambda$ aur $\mu$ (jahan $\mu = G$ shear modulus hai). Iska clean form yaad rakho: $\sigma_{ij} = \lambda\,\varepsilon_{kk}\delta_{ij} + 2\mu\,\varepsilon_{ij}$ . Yahaan $\varepsilon_{kk}$ trace hai jo volume change batata hai.

Engineering form mein hum likhte hain $\varepsilon_{xx} = \frac1E[\sigma_{xx} - \nu(\sigma_{yy}+\sigma_{zz})]$ — ye $-\nu(...)$ wala part hi Poisson coupling hai. Ek trap yaad rakho: shear mein engineering shear $\gamma_{xy} = 2\varepsilon_{xy}$ use karna, warna factor-of-2 ki galti ho jaati hai aur answer double/half ho jaata hai.

Ye kyun important hai spacecraft ke liye? Jab pressure tank ko pressurize karte ho,

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

WHY do we need a tensor at all?

Deriving strain from displacement (first principles)

Counting the constants: 81 → 21 → 2

The isotropic law (the workhorse)

Inverting to strain-from-stress (engineering form)

Voigt (matrix) notation — how engineers actually compute

Worked examples

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Spacecraft Structures & Systems Engineering

Connections