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

This page is the exercise ground for Hooke's law in 3D — generalized stress-strain (tensor). We take the isotropic law and drive it through every kind of loading a spacecraft part can feel — one pull, all-round squeeze, pure twist, two directions at once, the tricky sign cases, and the degenerate limits where the formula almost breaks.

Before line one, here are the only three tools we use, stated in plain words:

Meaning of the sign convention, always: positive stress = pull (tension), negative stress = push (compression); positive strain = it got longer in that direction, negative strain = it got shorter.

The scenario matrix

Every problem below is one cell of this grid. The columns are what kind of loading; the rows are what makes it tricky. We will tick off every cell.

Cell	Loading type	The twist that must be shown
A	Single pull (uniaxial, $σ > 0$ )	Recover 1D + see sideways shrink
B	Single push (uniaxial, $σ < 0$ )	Negative sign flows through cleanly
C	All-round squeeze (hydrostatic)	Volume-only, defines $K$
D	Pure twist (shear only)	No volume change, only $G$ enters
E	Two directions at once (biaxial)	Poisson coupling between the two
F	Mixed tension + compression	Signs partly cancel inside the bracket
G	Degenerate limit $ν \to \frac{1}{2}$	Incompressible — $K \to \infty$
H	Degenerate limit $ν \to 0$	No coupling — axes go independent
I	Real-world word problem	Spacecraft tank, numbers with units
J	Exam twist	"Zero strain in $y$ " (a constraint, not a stress)

Figures accompany the geometric cells (A/B, C, D).

Worked examples

Worked example Cell A — Single pull (uniaxial tension)

An aluminium strut ( $E = 70 GPa$ , $ν = 0.33$ ) is pulled with $σ_{xx} = + 140 MPa$ ; nothing on the other faces.

Forecast: guess the sign of $ε_{y y}$ before reading on. Does the bar get fatter or thinner sideways when you pull it?

Along the pull: $ε_{xx} = \frac{1}{E} [σ_{xx} - ν (0 + 0)] = \frac{140}{70000} = 2.0 \times 1 0^{- 3}$ . Why this step? With only one stress, the Poisson bracket empties out and 3D collapses to plain $σ = E ε$ .
Sideways: $ε_{y y} = ε_{z z} = \frac{1}{E} [0 - ν σ_{xx}] = \frac{- 0.33 \times 140}{70000} = - 6.6 \times 1 0^{- 4}$ . Why this step? The $- ν σ_{xx}$ term is the Poisson effect — pulling in $x$ steals width from $y$ and $z$ . Negative = thinner. See
— the red bar stretches, the mint arrows show it pinching in.

Verify: $ε_{xx} / ε_{y y} = 2.0 \times 1 0^{- 3} / (- 6.6 \times 1 0^{- 4}) = - 1/0.33 = - 1/ ν$ . Exactly what Poisson's ratio means: sideways strain is $- ν$ times the axial strain. Units: strain is dimensionless (MPa/MPa). ✔

Worked example Cell B — Single push (uniaxial compression)

Same strut, now shoved: $σ_{xx} = - 140 MPa$ .

Forecast: which way do the sideways faces move now — in or out?

Along: $ε_{xx} = \frac{- 140}{70000} = - 2.0 \times 1 0^{- 3}$ (it shortens — negative, as expected). Why this step? Same formula; the minus sign of the stress just rides straight through. Nothing special happens for compression — the linear law is symmetric.
Sideways: $ε_{y y} = \frac{1}{E} [0 - ν (- 140)] = + 6.6 \times 1 0^{- 4}$ (positive → it bulges out). Why this step? Now the Poisson term is $- ν \times (negative) = +$ . Squeezing lengthwise fattens the sides — exactly the foam-block picture.

Verify: every number is the Cell-A answer with its sign flipped. Linear elasticity means "push = negative pull", and the ratio $- ν$ is preserved: $6.6 \times 1 0^{- 4} / (- 2.0 \times 1 0^{- 3}) = - 0.33 = - ν$ . ✔

Worked example Cell C — All-round squeeze (hydrostatic)

The strut is dropped to the bottom of a pressure chamber: $σ_{xx} = σ_{y y} = σ_{z z} = - p$ , $p = 140 MPa$ . No shear.

Forecast: with equal push on all six faces, is there any change of shape, or only of size?

One normal strain: $ε_{xx} = \frac{1}{E} [- p - ν (- p - p)] = \frac{- p ( 1 - 2 ν )}{E}$ . Why this step? Each face feels its own $- p$ , then gets partly pushed back out by the Poisson bulge of the other two, giving the factor $(1 - 2 ν)$ .
By symmetry $ε_{y y} = ε_{z z} = ε_{xx}$ , so no face is different — shape is unchanged, only volume. See
: the cube shrinks uniformly, staying a cube.
Volume change: $ε_{k k} = 3 ε_{xx} = \frac{- 3 p ( 1 - 2 ν )}{E}$ . Why this step? Trace of strain = fractional volume change; here it's the sum of three equal terms.
Numbers ( $ν = 0.33$ ): $ε_{k k} = \frac{- 3 \times 140 \times ( 1 - 0.66 )}{70000} = \frac{- 3 \times 140 \times 0.34}{70000} = - 2.04 \times 1 0^{- 3}$ .

Verify: the bulk modulus is $K = - p / ε_{k k} = \frac{140}{2.04 \times 1 0 ^{- 3}} = 68.6 GPa$ , which must equal $\frac{E}{3 ( 1 - 2 ν )} = \frac{70000}{3 \times 0.34} = 68.6 GPa$ . ✔

Worked example Cell D — Pure twist (shear only)

A spacecraft drive shaft carries $τ_{x y} = τ_{0} = 50 MPa$ ; all normal stresses zero. Same material.

Forecast: will this loading change the shaft's volume at all?

Shear strain: $γ_{x y} = \frac{τ _{0}}{G}$ , with $G = \frac{E}{2 ( 1 + ν )} = \frac{70000}{2 \times 1.33} = 26.3 GPa$ . Why this step? Shear rows of the compliance matrix are decoupled from the normal rows — a shear stress can produce only shear strain.
Number: $γ_{x y} = \frac{50}{26315} = 1.90 \times 1 0^{- 3} rad$ .
Volume? All $ε_{xx} = ε_{y y} = ε_{z z} = 0$ , so $ε_{k k} = 0$ : pure shape-change, zero volume change. See
— the pastel square racks over into a rhombus while keeping the same area.

Verify: rewrite $γ$ using $E, ν$ : $γ_{x y} = \frac{2 ( 1 + ν )}{E} τ_{0} = \frac{2 \times 1.33 \times 50}{70000} = 1.90 \times 1 0^{- 3}$ . Same value — the compliance shear entry $2 (1 + ν) / E$ really is $1/ G$ . ✔

Worked example Cell E — Two directions at once (biaxial)

A flat panel is pulled in-plane: $σ_{xx} = + 120 MPa$ , $σ_{y y} = + 60 MPa$ , $σ_{z z} = 0$ (plane stress). Same material.

Forecast: will $ε_{z z}$ (the thickness strain) be zero, positive, or negative?

$x$ -strain: $ε_{xx} = \frac{1}{E} [120 - ν (60 + 0)] = \frac{120 - 0.33 \times 60}{70000} = \frac{100.2}{70000} = 1.431 \times 1 0^{- 3}$ . Why this step? The $y$ -pull steals a little of the $x$ -stretch via Poisson.
$y$ -strain: $ε_{y y} = \frac{1}{E} [60 - ν (120 + 0)] = \frac{60 - 39.6}{70000} = \frac{20.4}{70000} = 2.914 \times 1 0^{- 4}$ .
Thickness strain: $ε_{z z} = \frac{1}{E} [0 - ν (120 + 60)] = \frac{- 0.33 \times 180}{70000} = - 8.486 \times 1 0^{- 4}$ . Why this step? Even though nothing pushes on the flat faces, both in-plane pulls squeeze the thickness down — that's why thin sheets get thinner when stretched biaxially.

Verify: $ε_{z z}$ must equal $- \frac{ν}{E} (σ_{xx} + σ_{y y}) = - \frac{0.33}{70000} \times 180 = - 8.486 \times 1 0^{- 4}$ . ✔ And it's negative, as forecast.

Worked example Cell F — Mixed tension + compression

A bracket sees $σ_{xx} = + 90 MPa$ (pull) but $σ_{y y} = - 90 MPa$ (push), $σ_{z z} = 0$ .

Forecast: in the $x$ -strain, does the Poisson term add to or subtract from the direct pull?

$x$ -strain: $ε_{xx} = \frac{1}{E} [90 - ν (- 90 + 0)] = \frac{90 + 0.33 \times 90}{70000} = \frac{119.7}{70000} = 1.710 \times 1 0^{- 3}$ . Why this step? Here the neighbour stress is negative, so $- ν σ_{y y} = + 29.7$ adds to the stretch: pulling $x$ while squeezing $y$ elongates $x$ even more. Signs matter enormously.
$y$ -strain: $ε_{y y} = \frac{1}{E} [- 90 - ν (90 + 0)] = \frac{- 90 - 29.7}{70000} = \frac{- 119.7}{70000} = - 1.710 \times 1 0^{- 3}$ .
Volume: $ε_{z z} = \frac{1}{E} [0 - ν (90 - 90)] = 0$ , so $ε_{k k} = ε_{xx} + ε_{y y} + 0 = 0$ . Why this step? Equal-and-opposite normal stresses is a pure shear state in disguise — no volume change, pure distortion.

Verify: $ε_{xx} = - ε_{y y}$ and $ε_{k k} = 0$ , the signature of a shape-only deformation. Check $ε_{xx} = \frac{90 ( 1 + ν )}{E} = \frac{90 \times 1.33}{70000} = 1.710 \times 1 0^{- 3}$ . ✔

Worked example Cell G — Degenerate limit

$ν \to \frac{1}{2}$ (incompressible) Take the Cell-C hydrostatic result and let $ν \to \frac{1}{2}$ (rubber-like seal material), $p = 140 MPa$ .

Forecast: what happens to the volume change as $ν$ climbs toward $0.5$ ?

Volume change: $ε_{k k} = \frac{- 3 p ( 1 - 2 ν )}{E}$ . As $ν \to \frac{1}{2}$ , the factor $(1 - 2 ν) \to 0$ , so $ε_{k k} \to 0$ . Why this step? The material refuses to change volume no matter how hard you squeeze — it just redistributes shape.
Bulk modulus: $K = \frac{E}{3 ( 1 - 2 ν )} \to \infty$ as $ν \to \frac{1}{2}$ . Why this step? Infinite resistance to volume change = incompressible. Physically $ν$ can never exceed $\frac{1}{2}$ , or $K$ would go negative (the material would expand when squeezed — unstable).

Verify (numeric): at $ν = 0.499$ , $K = \frac{70000}{3 ( 1 - 0.998 )} = \frac{70000}{0.006} = 1.167 \times 1 0^{7} MPa$ — enormous, heading to infinity. ✔

Worked example Cell I — Real-world word problem (spacecraft tank)

A thin-walled cylindrical propellant tank (thin-walled vessel): radius $r = 1.0 m$ , wall $t = 5 mm$ , internal pressure $p = 2.0 MPa$ . Aluminium, $E = 70 GPa$ , $ν = 0.33$ . Find the axial strain $ε_{z}$ (needed to check bolted flanges don't leak).

Forecast: hoop stress is twice the axial stress. Will Poisson coupling raise or lower the axial strain compared to axial stress alone?

Hoop stress: $σ_{θ} = \frac{p r}{t} = \frac{2.0 \times 1.0}{0.005} = 400 MPa$ . Why this step? Standard thin-wall formula; the hoop direction carries the most load.
Axial stress: $σ_{z} = \frac{p r}{2 t} = \frac{2.0 \times 1.0}{2 \times 0.005} = 200 MPa$ . Radial $\approx 0$ (plane stress).
Axial strain: $ε_{z} = \frac{1}{E} [σ_{z} - ν (σ_{θ} + 0)] = \frac{200 - 0.33 \times 400}{70000} = \frac{68}{70000} = 9.714 \times 1 0^{- 4}$ . Why this step? The big hoop stress pulls the wall sideways and (via Poisson) shortens the tank axially, cutting the axial strain almost in half.

Verify: using the tidy form $ε_{z} = \frac{p r}{2 tE} (1 - 2 ν) = \frac{2 \times 1 \times ( 1 - 0.66 )}{2 \times 0.005 \times 70000} = \frac{0.68}{700} = 9.714 \times 1 0^{- 4}$ . Matches step 3. ✔

Worked example Cell J — Exam twist (a constraint, not a stress)

A block is pulled with $σ_{xx} = + 150 MPa$ but is rigidly held in the $y$ -direction so that $ε_{y y} = 0$ (it cannot contract in $y$ ). The $z$ -face is free ( $σ_{z z} = 0$ ). Same material. Find the reaction stress $σ_{y y}$ and the resulting $ε_{xx}$ .

Forecast: to stop the block shrinking in $y$ , must the wall push (compress) or pull (tension) on it?

Impose the constraint on the $y$ -equation: $ε_{y y} = \frac{1}{E} [σ_{y y} - ν (σ_{xx} + σ_{z z})] = 0$ . Why this step? We do not know $σ_{y y}$ ; instead we know the strain there is zero — the wall supplies whatever stress is needed. This is the exam trap: a fixed strain becomes an unknown stress.
Solve for $σ_{y y}$ : $σ_{y y} = ν (σ_{xx} + 0) = 0.33 \times 150 = + 49.5 MPa$ (tension — the wall pulls back to hold it). Why this step? Without the wall the block would contract in $y$ ; the wall applies tension to cancel that contraction exactly.
Now the $x$ -strain: $ε_{xx} = \frac{1}{E} [σ_{xx} - ν (σ_{y y} + σ_{z z})] = \frac{150 - 0.33 \times 49.5}{70000} = \frac{133.665}{70000} = 1.909 \times 1 0^{- 3}$ . Why this step? The now-nonzero $σ_{y y}$ feeds back via Poisson and slightly stiffens the $x$ -response compared to free tension.

Verify: plug $σ_{y y} = 49.5$ back into the $y$ -equation: $\frac{1}{E} [49.5 - 0.33 (150 + 0)] = \frac{1}{E} [49.5 - 49.5] = 0$ . Constraint satisfied. ✔

Recall Which cell am I in?

Only one stress given, rest zero ::: uniaxial (Cell A/B) — collapses to $σ = E ε$ with Poisson on the sides. Equal normal stresses on all faces ::: hydrostatic (Cell C) — volume only, defines $K$ . Only a shear stress given ::: pure shear (Cell D) — use $γ = τ / G$ , no volume change. A strain is fixed to a value (e.g. $ε_{y y} = 0$ ) instead of a stress ::: constraint problem (Cell J) — solve for the reaction stress first. $ν \to \frac{1}{2}$ ::: incompressible, $K \to \infty$ . $ν \to 0$ ::: fully uncoupled, every axis an independent 1D spring.

Related tools once you've mastered these cases: Elastic strain energy density (energy stored per case), Von Mises yield criterion (when do these stresses cause yielding), and Finite Element Method — stiffness matrices (how a computer solves millions of these at once).