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

Before we compute anything, let us fix every word and symbol so none is used before it is earned.

Definition The symbols you will keep meeting

$σ_{ij}$ = stress: force per unit area on a tiny cube face. First index = which face (its outward normal), second = which way the force pushes. Units: pascals (Pa = N/m²).
When the two indices are equal ( $σ_{xx}$ ) it is a normal (push/pull) stress; when they differ it is a shear. Shear stresses are so common they get their own letter: $τ_{x y} \equiv σ_{x y}$ (just a rename — a force on the $x$ -face pointing along $y$ ).
$ε_{ij}$ = strain: fractional deformation, a pure number (metre-per-metre).
$γ_{x y}$ = engineering shear strain: defined as $γ_{x y} = 2 ε_{x y}$ . It measures the total angle change of a right corner; the factor of $2$ is because both the $x$ - and $y$ -sides tilt, and engineers add both tilts into one number.
$δ_{ij}$ = Kronecker delta: a bookkeeping switch that equals $1$ when $i = j$ and $0$ when $i \neq = j$ . It is the "identity" — it turns a formula ON only for matching indices (the diagonal / normal directions) and OFF for the off-diagonal (shear) ones.
$E$ = Young's modulus: how much stress it takes to stretch by a given fraction (Pa). Big $E$ = stiff.
$ν$ (nu) = Poisson's ratio: pull it long one way, this number tells how much it thins sideways. A pure ratio between $0$ and $\frac{1}{2}$ for ordinary solids.
$λ, μ$ = Lamé parameters: the two numbers that fully describe an isotropic material. $μ \equiv G$ is the shear modulus (resistance to twist); $λ$ links a volume change to normal stress. They are just repackagings of $E, ν$ (formulas below). Repeated index (like $σ_{k k}$ ) means add up $σ_{11} + σ_{22} + σ_{33}$ — the trace.

Throughout, when a material is named, use these values (typical aluminium alloy for spacecraft): $E = 70 GPa, ν = 0.33, G = \frac{E}{2 ( 1 + ν )} \approx 26.3 GPa .$

L1 — Recognition

1.1 In the symbol $σ_{23}$ , which face of the cube is loaded and in what direction does the force point? Is $σ_{23}$ equal to $σ_{32}$ , and why?

1.2 A material has $ε_{11} = ε_{22} = ε_{33} = 0.001$ and all shear strains zero. Write down the volumetric strain $ε_{k k}$ and say in words what it means.

1.3 Using the compliance matrix $[S]$ reproduced above, which single entry tells you the Poisson coupling between a stress in $x$ and the strain it causes in $y$ ? State its value in terms of $E, ν$ .

Recall Solutions — L1

1.1 First index $= 2$ : the face whose outward normal points along the $2$ -axis (the $y$ -face). Second index $= 3$ : the force on that face points along the $3$ -axis ( $z$ ). So $σ_{23}$ is a shear on the $y$ -face pushing in $z$ . Yes, $σ_{23} = σ_{32}$ : if they differed the tiny cube would feel a net twisting moment and spin up infinitely fast — moment balance forbids it, so the stress tensor is symmetric.

1.2 $ε_{k k} = ε_{11} + ε_{22} + ε_{33} = 0.001 + 0.001 + 0.001 = 0.003$ . It is the fractional change in volume: a cube of this material swells by $0.3%$ in volume.

1.3 The off-diagonal entry in the top-left $3 \times 3$ block: $S_{12} = - \frac{ν}{E}$ . So $ε_{xx}$ picks up $- \frac{ν}{E} σ_{y y}$ — the Poisson pull-in.

L2 — Application

2.1 A bar is loaded uniaxially with $σ_{xx} = 140$ MPa, everything else zero ( $E = 70$ GPa, $ν = 0.33$ ). Find $ε_{xx}$ , $ε_{y y}$ , $ε_{z z}$ .

2.2 A block is under equal triaxial tension $σ_{xx} = σ_{y y} = σ_{z z} = 100$ MPa, no shear. Find $ε_{xx}$ and the volumetric strain $ε_{k k}$ .

2.3 Pure shear $τ_{x y} = 60$ MPa acts alone (recall $τ_{x y} \equiv σ_{x y}$ , and the engineering shear strain is $γ_{x y} = 2 ε_{x y} = τ_{x y} / G$ ). Find $γ_{x y}$ . Confirm no normal strains appear.

Recall Solutions — L2

2.1 Only $σ_{xx}$ is nonzero, so $σ_{k k} = σ_{xx}$ . $ε_{xx} = \frac{1}{E} [σ_{xx} - ν (0 + 0)] = \frac{140 \times 1 0 ^{6}}{70 \times 1 0 ^{9}} = 2.0 \times 1 0^{- 3} .$ WHAT we did: dropped the two zero cross-stresses. WHY: uniaxial means the $y, z$ stresses vanish. $ε_{y y} = ε_{z z} = \frac{1}{E} [0 - ν σ_{xx}] = - 0.33 \times 2.0 \times 1 0^{- 3} = - 6.6 \times 1 0^{- 4} .$ The bar stretches $0.2%$ along $x$ and thins $0.066%$ each transverse way — the Poisson effect.

2.2 With all three normal stresses equal to $σ_{0} = 100$ MPa, $σ_{k k} = 3 σ_{0}$ .

=\frac{100\times10^6\,(1-0.66)}{70\times10^9}=4.857\times10^{-4}.$$ $$\varepsilon_{kk}=3\varepsilon_{xx}=1.457\times10^{-3}.$$ Note it is positive (tension swells volume) and small because $(1-2\nu)$ is small. **2.3** Shear is decoupled, so use $\gamma_{xy}=\tau_{xy}/G$ with $G=\dfrac{E}{2(1+\nu)}=\dfrac{70\times10^9}{2(1.33)}=26.316\times10^9$ Pa: $$\gamma_{xy}=\frac{60\times10^6}{26.316\times10^9}=2.28\times10^{-3}.$$ Normal strains come only from normal stresses in the compliance matrix; all normal stresses are zero here, so $\varepsilon_{xx}=\varepsilon_{yy}=\varepsilon_{zz}=0$. Pure distortion, no volume change.

L3 — Analysis

3.1 (Plane stress) A thin flat panel has $σ_{z z} = 0$ (nothing pushes through its thickness). Given $σ_{xx} = 200$ MPa, $σ_{y y} = 80$ MPa, $τ_{x y} = 0$ , find $ε_{z z}$ . Explain why it is nonzero even though $σ_{z z} = 0$ .

3.2 (Plane strain) A long tunnel-like body is fully constrained along $z$ so $ε_{z z} = 0$ . With $σ_{xx} = 200$ MPa, $σ_{y y} = 80$ MPa given, find the induced $σ_{z z}$ that the constraint must supply.

3.3 For the plane-strain body above, using the $σ_{z z}$ you found, compute $ε_{xx}$ .

The figure below contrasts the two cases. Left (plane stress): a thin panel is pulled by in-plane forces (black arrows) while its two flat faces are free; the red arrows show it thinning through the thickness — $σ_{z z} = 0$ but $ε_{z z} \neq = 0$ . Right (plane strain): the same in-plane pull, but the top and bottom faces are clamped (red hatching) so the thickness cannot change — $ε_{z z} = 0$ , and the clamps must supply a hidden $σ_{z z} \neq = 0$ . The accent colour marks, in each panel, the thing that is not zero.

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

Recall Solutions — L3

3.1 Plane stress: $σ_{z z} = 0$ , so $σ_{k k} = σ_{xx} + σ_{y y} = 280$ MPa.

=\frac{1}{70\times10^9}[0-0.33(280\times10^6)]=-1.32\times10^{-3}.$$ *WHY nonzero:* no *force* pushes through the thickness, but the in-plane pulls squeeze the panel thinner through the Poisson effect. Zero stress $\ne$ zero strain. **3.2** Plane strain forces $\varepsilon_{zz}=0$. Set the $z$-strain equation to zero: $$0=\frac1E[\sigma_{zz}-\nu(\sigma_{xx}+\sigma_{yy})]\ \Rightarrow\ \sigma_{zz}=\nu(\sigma_{xx}+\sigma_{yy})=0.33(280\times10^6)=92.4\ \text{MPa}.$$ *WHY:* to stop the natural Poisson thinning, the walls must *push back* along $z$ with exactly this stress. The constraint manufactures a real stress out of nothing but geometry. **3.3** Now $\sigma_{kk}=\sigma_{xx}+\sigma_{yy}+\sigma_{zz}=200+80+92.4=372.4$ MPa. The Poisson pull-in on $\varepsilon_{xx}$ comes from $\sigma_{yy}+\sigma_{zz}=80+92.4=172.4$ MPa: $$\varepsilon_{xx}=\frac1E[\sigma_{xx}-\nu(\sigma_{yy}+\sigma_{zz})] =\frac{[200-0.33(172.4)]\times10^6}{70\times10^9}=2.04\times10^{-3}.$$ Compare: the induced $\sigma_{zz}$ adds extra Poisson pull-in versus a free panel, but $\sigma_{xx}$ still dominates, so $\varepsilon_{xx}$ stays positive (the body stretches along $x$).

L4 — Synthesis

4.1 (Pressure-vessel wall) A thin cylindrical propellant tank has hoop stress $σ_{θ} = \frac{p r}{t}$ , axial stress $σ_{z} = \frac{p r}{2 t}$ , radial $\approx 0$ (plane stress). With $p = 2$ MPa, $r = 0.5$ m, $t = 4$ mm, $E = 70$ GPa, $ν = 0.33$ , find $σ_{θ}$ , $σ_{z}$ , then the axial strain $ε_{z}$ and hoop strain $ε_{θ}$ . See Thin-walled pressure vessels.

4.2 For the same tank, compute the von Mises equivalent stress $σ_{v} = σ_{θ}^{2} - σ_{θ} σ_{z} + σ_{z}^{2}$ (plane-stress form) and compare to a yield strength of $270$ MPa — does the wall yield? See Von Mises yield criterion.

4.3 Derive the bulk modulus $K$ purely from the isotropic law by loading with pure hydrostatic pressure $σ_{xx} = σ_{y y} = σ_{z z} = - p$ , then plug $E = 70$ GPa, $ν = 0.33$ for a number.

Recall Solutions — L4

4.1 Stresses first:

\sigma_z=\tfrac12\sigma_\theta=125\ \text{MPa}.$$ Axial strain (radial stress $=0$, so only $\sigma_\theta$ is the Poisson partner): $$\varepsilon_z=\frac1E[\sigma_z-\nu\sigma_\theta] =\frac{[125-0.33(250)]\times10^6}{70\times10^9}=6.07\times10^{-4}.$$ Hoop strain (its Poisson partner is $\sigma_z$): $$\varepsilon_\theta=\frac1E[\sigma_\theta-\nu\sigma_z] =\frac{[250-0.33(125)]\times10^6}{70\times10^9}=2.98\times10^{-3}.$$ *WHY the coupling:* in the thin wall the hoop and axial directions are perpendicular, so each acts as the *other's* Poisson partner. The big hoop stress tries to thin the wall axially (reducing $\varepsilon_z$ via the $-\nu\sigma_\theta$ term), while the smaller axial stress only mildly reduces the hoop stretch (the $-\nu\sigma_z$ term). That is why $\varepsilon_\theta$ ends up much larger than $\varepsilon_z$ — and why sizing bolted flanges must account for this coupled thinning. **4.2** $\sigma_v=\sqrt{250^2-250\cdot125+125^2}=\sqrt{62500-31250+15625}=\sqrt{46875}=216.5$ MPa. Since $216.5 < 270$ MPa, the wall does **not** yield — but the margin is only about $20\%$, tight for flight hardware. **4.3** With $\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=-p$: each normal strain is $\varepsilon_{xx}=\frac1E[-p-\nu(-2p)]=\frac{-p(1-2\nu)}{E}$. Then $\varepsilon_{kk}=3\varepsilon_{xx}=\frac{-3p(1-2\nu)}{E}$. Bulk modulus is $K\equiv -p/\varepsilon_{kk}$: $$K=\frac{E}{3(1-2\nu)}=\frac{70\times10^9}{3(1-0.66)}=\frac{70\times10^9}{1.02}=68.6\ \text{GPa}.$$

L5 — Mastery

5.1 Starting from the strain-from-stress law $ε_{ij} = \frac{1 + ν}{E} σ_{ij} - \frac{ν}{E} σ_{k k} δ_{ij}$ , derive the shear relation $γ_{x y} = τ_{x y} / G$ and thereby prove $G = \frac{E}{2 ( 1 + ν )}$ from first principles.

5.2 (Energy) The elastic strain-energy density is $W = \frac{1}{2} σ_{ij} ε_{ij}$ (see Elastic strain energy density). For the uniaxial bar of 2.1 ( $σ_{xx} = 140$ MPa, others zero), show $W = \frac{1}{2} σ_{xx} ε_{xx}$ and evaluate it in J/m³.

5.3 (Incompressibility limit) Show algebraically that as $ν \to \frac{1}{2}$ , the bulk modulus $K \to \infty$ , and explain physically why $ν > \frac{1}{2}$ is forbidden. Then compute how much $K$ changes if a rubber-like material has $ν = 0.499$ vs $ν = 0.4999$ (take $E = 10$ MPa).

Recall Solutions — L5

5.1 For the off-diagonal component $i \neq = j$ , the Kronecker delta $δ_{ij} = 0$ , so the whole trace term drops: $ε_{x y} = \frac{1 + ν}{E} σ_{x y} .$ Engineering shear is $γ_{x y} = 2 ε_{x y} = \frac{2 ( 1 + ν )}{E} τ_{x y}$ . By definition $γ_{x y} = τ_{x y} / G$ , so matching coefficients: $\frac{1}{G} = \frac{2 ( 1 + ν )}{E} \Rightarrow G = \frac{E}{2 ( 1 + ν )} .$ WHAT/WHY: the delta term is the only thing linking shear to the volumetric part; killing it ( $i \neq = j$ ) isolates pure shear, and the factor-of-2 comes straight from the $γ = 2 ε$ convention.

5.2 With only $σ_{xx}$ nonzero, the sum $W = \frac{1}{2} σ_{ij} ε_{ij}$ has just one surviving term (all other $σ_{ij} = 0$ ):

=\tfrac12\,\frac{(140\times10^6)^2}{70\times10^9}=1.4\times10^{5}\ \text{J/m}^3.$$ (Cross terms like $\sigma_{yy}\varepsilon_{yy}$ vanish because $\sigma_{yy}=0$, even though $\varepsilon_{yy}\ne0$ — energy needs *both* factors.) **5.3** $K=\dfrac{E}{3(1-2\nu)}$. As $\nu\to\tfrac12$, the denominator $(1-2\nu)\to0^+$, so $K\to+\infty$: it takes infinite pressure to change the volume — the material is **incompressible**. If $\nu>\tfrac12$ then $(1-2\nu)<0$ and $K<0$, meaning squeezing would make it *expand*: a spontaneous instability, physically impossible for a passive solid. Hence $\nu\le\tfrac12$. Numbers with $E=10$ MPa: $$K(0.499)=\frac{10^7}{3(1-0.998)}=\frac{10^7}{0.006}=1.667\times10^{9}\ \text{Pa},$$ $$K(0.4999)=\frac{10^7}{3(1-0.9998)}=\frac{10^7}{0.0006}=1.667\times10^{10}\ \text{Pa}.$$ A tenfold approach of $\nu$ toward $\tfrac12$ makes $K$ ten times larger — the blow-up is steep.

Recall One-line self-check before you leave

Which convention pitfall bit you most — face-vs-force indices, $γ$ -vs- $ε$ shear, or plane-stress-vs-plane-strain? ::: All three are "you fixed the wrong one of a symmetric pair" — read the physics first, then choose which member is zero.