Worked examples — Composite materials — fiber-matrix, ply properties, laminate theory

This page is a drill deck. The parent note built the machinery; here we run it against every kind of input the topic can hand you. Before each worked example, you make a forecast — a guess. Guessing first is how the ideas stick.

Everything below only uses tools the parent earned: the Rule of Mixtures ($E_1=E_f{V}_f+E_m{V}_m$), its inverse cousin for transverse and shear, the ply stiffness matrix $[Q]$, and the angle transformation $[\bar{Q}]$. Nothing new is smuggled in.

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The scenario matrix

Think of every problem this topic throws as living in one cell of this table. Our job: hit all of them.

#	Case class	What makes it tricky	Covered by
A	Longitudinal $E_1$ (parallel springs)	straightforward, but the baseline	Ex 1
B	Transverse $E_2$ / shear $G_{12}$ (series springs)	matrix dominates — counter-intuitive	Ex 2
C	Degenerate: $V_f=0$ and $V_f=1$	limiting values — does the formula reduce to pure matrix / pure fiber?	Ex 3
D	Building $[Q]$ from $E_1,E_2,G_{12},{\nu }_{12}$	Poisson coupling, the $1-{\nu }_{12}{\nu }_{21}$ denominator	Ex 4
E	Off-axis at $\theta =0^{\circ }$ and $90^{\circ }$	check the transform doesn't lie at the trivial angles	Ex 5
F	Off-axis at $\theta =45^{\circ }$	the extension–shear coupling appears — the whole point	Ex 6
G	Real-world word problem: mass budget	choosing composite vs aluminium by specific stiffness	Ex 7
H	Exam twist: work backwards	given $E_1$, find the required $V_f$	Ex 8

Two golden rules travel with us:

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Ex 1 — Case A: Longitudinal modulus (parallel springs)

Forecast: the fibers are ~77× stiffer than the matrix and fill 60% of the volume. Guess: will $E_1$ land closer to $230$ or closer to $3$? (Closer to fiber — the stiff phase wins when strains are shared.)

1. Get the matrix fraction. $V_m=1-V_f=1-0.60=0.40$. Why this step? Volume fractions must sum to 1 — every bit of the cube is either fiber or matrix, nothing else.
2. Add the stiffnesses weighted by volume (parallel springs — look at the two arrows on the left of the figure, both stretched by the same amount): $E_1=E_f{V}_f+E_m{V}_m=230(0.60)+3(0.40)=138+1.2=139.2\text{ GPa}.$ Why this step? Bonded fiber and matrix stretch equally, so each contributes stress $\propto$ its own stiffness. Total stress ÷ shared strain = weighted sum of moduli.

Verify: $E_1$ must sit between $E_m$ and $E_f$ ($3<139.2<230$ ✓) and lean toward the fiber because it dominates the sum. $139.2/230=0.605$, i.e. 60.5% of pure fiber — sensible, matching $V_f$. Units: GPa throughout ✓.

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Ex 2 — Case B: Transverse & shear (series springs)

Forecast: now the load crosses the soft matrix in series. Guess: does $E_2$ come out near $139$ GPa or a small number? (Small — the weakest link in a series controls it.)

1. Add the softnesses (reciprocals), because in series the two share stress and their strains add: $\frac{1}{E_2}=\frac{V_f}{E_f}+\frac{V_m}{E_m}=\frac{0.60}{230}+\frac{0.40}{3}=0.002609+0.13333=\mathrm{0.13594.}$ Why this step? In series (right side of the figure), the same stress passes through fiber then matrix; the total stretch is the sum of each layer's stretch → compliances add.
2. Invert to recover the modulus: $E_2=\frac{1}{0.13594}=7.36\text{ GPa}.$
3. Shear uses the same series rule (shear also flows through the soft matrix): $\frac{1}{G_{12}}=\frac{0.60}{9}+\frac{0.40}{1.1}=0.06667+0.36364=0.43030\Rightarrow G_{12}=2.32\text{ GPa}.$

Verify: $E_2=7.36$ GPa is barely twice the matrix modulus $3$ GPa — the fibers help transversely almost not at all. The anisotropy ratio $E_1/E_2=139.2/7.36\approx 18.9$, matching the parent's "$\approx 19$". Units consistent ✓. This is exactly why the parent warns never to treat composites as isotropic.

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Ex 3 — Case C: Degenerate limits $V_f=0$ and $V_f=1$

Forecast: a formula you can trust must give pure matrix properties at $V_f=0$ and pure fiber at $V_f=1$. Guess the four numbers before reading.

1. $V_f=0$ (all matrix), longitudinal: $E_1=230(0)+3(1)=3$ GPa. Why this step? No fiber left — the composite is the matrix, so $E_1$ must equal $E_m$.
2. $V_f=0$, transverse: $1/E_2=0/230+1/3=1/3\Rightarrow E_2=3$ GPa. Why this step? Same reasoning — both formulas collapse to $E_m$. Good, they agree at this endpoint.
3. $V_f=1$ (all fiber), longitudinal: $E_1=230(1)+3(0)=230$ GPa.
4. $V_f=1$, transverse: $1/E_2=1/230+0/3=1/230\Rightarrow E_2=230$ GPa. Why this step? Pure fiber has no direction preference in these idealised formulas, so $E_1=E_2=E_f$.

Verify: at both extremes $E_1=E_2$, exactly as a single material demands (anisotropy comes only from mixing). The endpoints are $E_m$ and $E_f$ — no blow-ups, no negative values, no division by zero anywhere in $0\le V_f\le 1$ ✓.

Recall Why can't

$V_f$ reach 1 in practice? Real fibers are round; packed cylinders leave gaps for matrix, so $V_f$ tops out near $\approx 0.65$–$0.75$. The formula still behaves at $V_f=1$, but manufacturing forbids it. ::: Round packing + need for the matrix to wet every fiber caps $V_f$ around 0.65–0.75.

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Ex 4 — Case D: Building the ply stiffness matrix $[Q]$

Forecast: the denominator $1-{\nu }_{12}{\nu }_{21}$ is only slightly less than 1, so guess: will $Q_{11}$ be close to $140$ GPa or very different? (Close — the Poisson correction is tiny.)

1. Minor Poisson ratio from symmetry. ${\nu }_{21}={\nu }_{12}{E}_2/E_1=0.30\times 10/140=0.02143$. Why this step? Stiffness matrices must be symmetric (energy can't depend on load order); that forces ${\nu }_{21}{E}_1={\nu }_{12}{E}_2$.
2. The shared denominator. $D=1-{\nu }_{12}{\nu }_{21}=1-0.30(0.02143)=1-0.006429=0.99357$. Why this step? Inverting the compliance matrix produces this determinant term; it "un-mixes" the Poisson coupling so stress↔strain stays consistent.
3. Fill the entries. $Q_{11}=\frac{E_1}{D}=\frac{140}{0.99357}=140.91\text{ GPa},Q_{22}=\frac{E_2}{D}=\frac{10}{0.99357}=10.065\text{ GPa},$ $Q_{12}=\frac{{\nu }_{12}{E}_2}{D}=\frac{0.30(10)}{0.99357}=3.019\text{ GPa},Q_{66}=G_{12}=5\text{ GPa}.$

Verify: $Q_{11}\gtrsim E_1$ (140.91 vs 140) — the Poisson coupling stiffens slightly, as predicted. Symmetry check: $Q_{12}$ computed either as ${\nu }_{12}{E}_2/D$ or ${\nu }_{21}{E}_1/D=0.02143(140)/0.99357=3.019$ GPa — identical ✓. Shear sits alone at $Q_{66}=G_{12}$ (no coupling to normal stress in material axes) ✓.

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Ex 5 — Case E: Transform at trivial angles $\theta =0^{\circ },90^{\circ }$

Forecast: at $0^{\circ }$ the fibers already line up with $x$, so nothing should change. At $90^{\circ }$ we've turned the ply a quarter turn — the stiff direction now points along $y$, so the $x$-stiffness should read the transverse value. Guess: ${\bar{Q}}_{11}(0^{\circ })=?$ and ${\bar{Q}}_{11}(90^{\circ })=?$

We use the standard result (equivalent to $[\bar{Q}]=[T{]}^{-1}[Q][T{]}^{-T}$) with $c=\cos \theta$, $s=\sin \theta$: ${\bar{Q}}_{11}=Q_{11}{c}^4+2(Q_{12}+2Q_{66})s^2{c}^2+Q_{22}{s}^4.$

1. $\theta =0^{\circ }$: $c=1,s=0$. Only the first term survives: ${\bar{Q}}_{11}=Q_{11}(1)=140.91$ GPa. Why this step? No rotation → material axes = laminate axes → the matrix is unchanged. If it changed, our transform would be broken.
2. $\theta =90^{\circ }$: $c=0,s=1$. Only the last term survives: ${\bar{Q}}_{11}=Q_{22}(1)=10.065$ GPa. Why this step? A quarter turn swaps the roles of directions 1 and 2, so the $x$-stiffness becomes the transverse stiffness $Q_{22}$ — the ply is now soft along $x$.

Verify: ${\bar{Q}}_{11}(0^{\circ })=Q_{11}$ and ${\bar{Q}}_{11}(90^{\circ })=Q_{22}$ — exactly the two endpoints of $[Q]$. The transform passes both trivial tests, so we can trust it at messy angles ✓.

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Ex 6 — Case F: Transform at $\theta =45^{\circ }$ — coupling appears

Forecast: at $45^{\circ }$ pulling along $x$ tugs on both fiber and transverse directions and twists the ply. So predict ${\bar{Q}}_{11}$ lands between $Q_{11}$ and $Q_{22}$, and ${\bar{Q}}_{16}$ becomes nonzero (it was zero in material axes).

At $\theta =45^{\circ }$: $c=s=\frac{1}{\sqrt{2}}$, so $c^2=s^2=\frac{1}{2}$, $c^4=s^4=\frac{1}{4}$, $s^2{c}^2=\frac{1}{4}$, and $c^2-s^2=0$.

1. Compute ${\bar{Q}}_{11}$: ${\bar{Q}}_{11}=Q_{11}\frac{1}{4}+2(Q_{12}+2Q_{66})\frac{1}{4}+Q_{22}\frac{1}{4}.$ $=\frac{1}{4}[140.91+2(3.019+2\cdot 5)+10.065]=\frac{1}{4}[140.91+26.038+10.065]=\frac{1}{4}(177.01)=44.25\text{ GPa}.$ Why this step? Weighting $Q_{11},Q_{22}$ and the shear-laden middle term equally (each $\times \frac{1}{4}$) mixes stiff, soft, and shear contributions — the ply is neither fully stiff nor fully soft along $x$.
2. Compute the coupling term ${\bar{Q}}_{16}=(Q_{11}-Q_{12}-2Q_{66})c^{3}s+(Q_{12}-Q_{22}+2Q_{66})c{s}^3$. At $45^{\circ }$, $c^{3}s=c{s}^3=\frac{1}{4}$: ${\bar{Q}}_{16}=\frac{1}{4}[(Q_{11}-Q_{12}-2Q_{66})+(Q_{12}-Q_{22}+2Q_{66})]=\frac{1}{4}(Q_{11}-Q_{22})=\frac{1}{4}(140.91-10.065)=32.71\text{ GPa}.$ Why this step? The $\pm 2Q_{66}$ and $\pm Q_{12}$ pieces cancel, leaving $\frac{1}{4}(Q_{11}-Q_{22})$ — the coupling is driven purely by how unequal the two direct stiffnesses are.

Verify: ${\bar{Q}}_{11}=44.25$ GPa sits between $Q_{22}=10.06$ and $Q_{11}=140.9$ ✓. ${\bar{Q}}_{16}=32.71\ne 0$ — pulling along $x$ now produces shear (the extension–shear coupling the parent flagged). This is why single off-axis plies are avoided and why balanced $\pm 45^{\circ }$ pairs are used (their ${\bar{Q}}_{16}$ terms cancel). Look at the tilted fibers in the figure — the pull no longer aligns with either weave direction.

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Ex 7 — Case G: Real-world word problem (mass budget)

Forecast: composite is stiffer and lighter — guess the specific-stiffness advantage is "a few ×", not 100×.

1. Specific stiffness of each (stiffness per unit density — the fair figure of merit when mass budgets rule): ${\left(\frac{E}{\rho }\right)}_{\text{comp}}=\frac{139.2}{1600}\times 10^6=87.0\times 10^3,{\left(\frac{E}{\rho }\right)}_{\text{Al}}=\frac{70}{2700}\times 10^6=25.9\times 10^3{\text{(SI: Pa⋅m}}^3\text{/kg).}$ Why this step? For strength-limited or stiffness-limited parts, the mass that meets a spec scales as $\rho /E$ — so maximising $E/\rho$ minimises mass.
2. Ratio: $87.0/25.9=3.36$. Why this step? Dividing gives the head-to-head advantage, cancelling the panel geometry that both designs share.

Verify: matches the parent's "3.3× better" specific stiffness ✓. Interpretation: a composite bench holding the same stiffness weighs roughly $1/3.36\approx 30\%$ of the aluminium one — exactly why launch-load-driven (quasi-static loads) structures go composite. Units cancel to a pure ratio ✓.

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Ex 8 — Case H: Exam twist — work backwards for $V_f$

Forecast: $150$ is comfortably above the $139.2$ we got at $V_f=0.6$, so guess $V_f$ needs to rise — but is it still below the $\approx 0.70$ ceiling?

1. Write the Rule of Mixtures as one unknown. With $V_m=1-V_f$: $E_1=E_f{V}_f+E_m(1-V_f)=E_m+(E_f-E_m)V_f.$ Why this step? Collapsing to a single $V_f$ turns the design goal into a linear equation we can invert.
2. Solve for $V_f$: $V_f=\frac{E_1-E_m}{E_f-E_m}=\frac{150-3}{230-3}=\frac{147}{227}=\mathrm{0.6476.}$ Why this step? Straight algebra — the same "between the endpoints" line, read the other way.

Verify: plug back: $E_1=3+227(0.6476)=3+147=150$ GPa ✓. And $V_f=0.648$ sits under the practical $\approx 0.65$–$0.70$ ceiling from Ex 3, so it is manufacturable — barely. If the target were $E_1=200$ GPa, we'd need $V_f=(200-3)/227=0.868>0.70$: impossible with this fiber, pushing you toward a stiffer fiber or a sandwich construction instead.

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Recall Which rule for which property?

$E_1$ uses direct Rule of Mixtures (parallel). ::: $E_1=E_f{V}_f+E_m{V}_m$. $E_2$, $G_{12}$ use the inverse (series) rule. ::: $1/E_2=V_f/E_f+V_m/E_m$. At $V_f\to 0$ both moduli approach ::: $E_m$ (pure matrix). The off-axis coupling term ${\bar{Q}}_{16}$ at $45^{\circ }$ equals ::: $\frac{1}{4}(Q_{11}-Q_{22})$.

These worked cases feed directly into laminate stacking and into FE models, where each ply's $[\bar{Q}]$ becomes an element property; thermal versions of the same algebra drive thermal-expansion design.