Worked examples — Polymer-matrix composites — CFRP, GFRP; ply lay-up, laminate theory

Everything here reuses just three tools from the parent, and we re-state each in plain words the first time it appears:

• Volume fraction $V_f$ = the slice of the material's volume that is fibre; $V_m=1-V_f$ is what's left for matrix. Think of a bundle of straws in glue: $V_f$ is how much of the cross-section is straw.
• Young's modulus $E$ = stiffness. Pull with stress $\sigma$ (force per area), get strain $\epsilon$ (fractional stretch); $E=\sigma /\epsilon$. Big $E$ = hard to stretch. See Stress, strain and Young's modulus.
• Iso-strain vs iso-stress = are the fibre and matrix stretching together (parallel, same strain) or one-behind-the-other (series, same stress)? This single question decides which formula you use.

Two more tools appear only in the bending examples (Ex 6–7). We define them there, right where they are needed, but here is the plain-words preview:

• $\bar{Q}$ (reduced stiffness) = a single number standing for "how stiff this ply is in the global loading direction," after we rotate its along-fibre/across-fibre stiffness to the laminate's own axes. Think of it as the ply's effective $E$ once it is glued into the stack at its chosen angle. (Full detail in Ex 6.)
• $D$ (bending stiffness) = how hard the whole laminate is to bend, as opposed to how hard it is to stretch. (Full detail in Ex 7.)

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The scenario matrix

Every problem this topic can pose falls into one of these cells. The worked examples afterwards are tagged with the cell they cover, so together they hit all of them.

#	Case class	The input that defines it	Which tool	Example
C1	Along fibres (longitudinal)	Load $\parallel$ fibres → iso-strain	$E_1=E_f{V}_f+E_m{V}_m$	Ex 1
C2	Across fibres (transverse)	Load $\perp$ fibres → iso-stress	$1/E_2=V_f/E_f+V_m/E_m$	Ex 2
C3	Zero / degenerate limits	$V_f=0$ (pure matrix) and $V_f=1$ (pure fibre)	both formulas at endpoints	Ex 3
C4	Low-$V_f$ limit	tiny $V_f$ — does adding fibre help?	derivative / sensitivity	Ex 4
C5	Load split & failure	how much force does the fibre carry?	force = stress × area	Ex 5
C6	Symmetry sign check	$+z$ vs $-z$ plies → does $B$ vanish?	$B_{ij}=\frac{1}{2}\sum \bar{Q}(z_k^2-z_{k-1}^2)$	Ex 6
C7	Bending, outer-ply weighting	same ply at surface vs centre	$D_{ij}=\frac{1}{3}\sum \bar{Q}(z_k^3-z_{k-1}^3)$	Ex 7
C8	Real-world word problem	aircraft skin, choose CFRP vs GFRP per kg	specific stiffness $E/\rho$	Ex 8
C9	Exam-style twist	"which formula?" trap + $V_f$ over-limit	recognise iso-stress vs iso-strain	Ex 9

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Ex 1 — Longitudinal modulus (cell C1)

Forecast: the answer will sit close to the big number (230) because the stiff fibres run the whole length and carry the pull. Guess: somewhere between 130 and 145 GPa. Write your guess down.

1. Identify the loading. Load runs along the fibres, so fibre and matrix stretch the same amount → same strain → iso-strain. Why this step? The whole choice of formula hinges on parallel-vs-series. Along the fibres they are in parallel (springs side by side).
2. Fill in volume fractions. $V_m=1-V_f=1-0.6=0.4$. Why? Fibre and matrix together make 100% of the volume; nothing else is present.
3. Apply the rule of mixtures. $E_1=E_f{V}_f+E_m{V}_m=230(0.6)+3(0.4)$ Why? Each phase contributes its stiffness weighted by how much of it there is — a straight weighted average.
4. Compute. $230(0.6)=138$, and $3(0.4)=1.2$, so $E_1=139.2$ GPa. Why? We just multiply out each weighted term and add; the fibre term (138) plainly dominates the matrix term (1.2), which is the physical point.

Verify: The result lies between $E_m=3$ and $E_f=230$ (a weighted average must). It sits near the fibre value because $V_f>0.5$. Units: GPa in, GPa out. ✓ Matches the forecast.

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Ex 2 — Transverse modulus (cell C2)

Forecast: load must pass through the soft matrix as if fibre and matrix were stacked in a line (series). A soft link in a chain dominates. Guess: much smaller than $E_1$ — maybe under 10 GPa.

1. Identify the loading. Load crosses the fibres → it goes fibre-then-matrix, in series → same stress, different strains → iso-stress. Why? In series, force passes through each link in turn, so each feels the same force-per-area.
2. Use the inverse (harmonic) rule. $\frac{1}{E_2}=\frac{V_f}{E_f}+\frac{V_m}{E_m}=\frac{0.6}{230}+\frac{0.4}{3}$ Why? In series, strains add, and strain $=\sigma /E$, so the reciprocals of stiffness add (like springs in series).
3. Compute the two terms. $0.6/230=0.002609$, $0.4/3=0.13333$. Sum $=0.13594$. Why? We evaluate each compliance term separately; note the second (matrix) term dwarfs the first — the soft matrix's high compliance ($1/E$) controls the total.
4. Invert. $E_2=1/0.13594=7.36$ GPa. Why? We solved for $1/E_2$, so to get $E_2$ we flip it back — the harmonic mean always lands near the smaller input, here $E_m$.

Verify: $E_2\ll E_1$ ($7.36$ vs $139.2$, ratio $\approx 19$). This is the anisotropy the parent note warned about, and it's why we lay up at angles. See Anisotropy and crystal directions. ✓

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Ex 3 — The two degenerate limits (cell C3)

Forecast: with no fibre, both $E_1$ and $E_2$ should equal $E_m$; with all fibre, both should equal $E_f$. If a formula fails a limit, it's wrong.

1. $V_f=0$ in the parallel rule. $E_1=E_f(0)+E_m(1)=E_m=3$ GPa. Why? No fibre present, so only matrix stiffness survives.
2. $V_f=0$ in the series rule. $1/E_2=0/E_f+1/E_m=1/E_m$, so $E_2=E_m=3$ GPa. Why? Same physical material both ways — a homogeneous block has one stiffness regardless of "direction."
3. $V_f=1$ in the parallel rule. $E_1=E_f(1)+E_m(0)=E_f=230$ GPa. Why? We set the fibre fraction to 100%; the matrix term multiplies by zero and drops out, leaving the pure-fibre value.
4. $V_f=1$ in the series rule. $1/E_2=1/E_f$, so $E_2=E_f=230$ GPa. Why? With no matrix in the series path, only the fibre's compliance remains; inverting gives $E_f$, and the block is uniform so both formulas must agree.

Verify: At each endpoint $E_1=E_2$ (no anisotropy when there's only one phase). Between the endpoints they split apart — that's exactly the fibre/matrix teamwork. Both formulas pass both limits. ✓

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Ex 4 — Does the first bit of fibre help most? (cell C4)

Forecast: along the fibres, even a little fibre should shoot the stiffness up (steep). Across the fibres, a bit of stiff fibre buried in a soft series-path barely helps (shallow). Guess: $E_1$ climbs fast, $E_2$ climbs slowly.

1. Slope of $E_1$. $E_1=E_f{V}_f+E_m(1-V_f)=E_m+(E_f-E_m)V_f$. This is a straight line, slope $=E_f-E_m=230-3=227$ GPa per unit $V_f$. Why derivative here? "Rate of gain" = how much output changes per unit input = the slope. For a straight line the slope is constant.
2. Slope of $E_2$ at $V_f=0$. From $1/E_2=V_f/E_f+(1-V_f)/E_m$, differentiate: $\frac{d}{d{V}_f}(1/E_2)=1/E_f-1/E_m$. Then $\frac{d{E}_2}{d{V}_f}=-E_2^2\left(\frac{1}{E_f}-\frac{1}{E_m}\right)$. At $V_f=0$, $E_2=E_m=3$: $\frac{d{E}_2}{d{V}_f}{|}_0=-3^2\left(\frac{1}{230}-\frac{1}{3}\right)=-9(0.004348-0.33333)=-9(-0.32899)=2.96\text{ GPa}.$ Why the chain rule? We know $1/E_2$ as a function of $V_f$; to get the slope of $E_2$ itself we differentiate $1/E_2$ and rescale by $-E_2^2$ (since $\frac{d}{dx}(1/y)=-y^{\prime }/y^2$).
3. Compare. $227$ GPa vs $2.96$ GPa per unit $V_f$ — the longitudinal stiffness grows about 77× faster. Why? We divide the two slopes ($227/2.96\approx 77$) to size the gap; the first sliver of fibre helps the along-fibre direction dramatically more.

Verify: Sign check — both slopes are positive (adding fibre never hurts stiffness). Magnitude check — $E_2$ barely moves because the soft matrix still forms an unbroken series path when fibres are sparse. This is the mathematical face of "a ply is useless off-axis." ✓

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Ex 5 — How the load splits, and who fails first (cell C5)

Forecast: fibres are ~77× stiffer than matrix and there's more of them ($0.6$ vs $0.4$). They should carry the overwhelming majority — guess >95%.

1. Same strain both phases (iso-strain). ${\epsilon }_f={\epsilon }_m=0.005$. Why? Bonded, pulled lengthwise → they stretch together (parallel), as in Ex 1.
2. Stress in each phase. $\sigma =E\epsilon$: ${\sigma }_f=230\times 0.005=1.15$ GPa; ${\sigma }_m=3\times 0.005=0.015$ GPa. Why? Each phase is elastic, so its stress is its own modulus times the shared strain.
3. Force fraction = stress × area fraction. With area fractions equal to volume fractions: $\text{fibre force fraction}=\frac{{\sigma }_f{V}_f}{{\sigma }_f{V}_f+{\sigma }_m{V}_m}=\frac{1.15(0.6)}{1.15(0.6)+0.015(0.4)}=\frac{0.690}{0.690+0.006}.$ Why? Force = stress × area; total is fibre-force plus matrix-force. Dividing gives the share.
4. Compute. $=0.690/0.696=0.9914\approx 99.1\%$. Why? We just evaluate the ratio; the matrix contribution $0.006$ is tiny next to $0.690$, so the fraction lands a hair under 1.

Verify: Sanity: the ply-average stress ${\sigma }_1=E_1\epsilon =139.2\times 0.005=0.696$ GPa — exactly the denominator's numerator sum $0.696$ GPa. ✓ Consistent. The matrix carries under 1%: it's a load-transfer medium, not a load-carrier. See Fibre–matrix interface and load transfer.

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Ex 6 — Symmetry sign check: does $B$ vanish? (cell C6)

Forecast: for a mirror-symmetric layup, every ply above the mid-plane has an identical twin below at $-z$. The $B$ integrand weights by $z^2$... but the twins sit at opposite signs of $z$; guess $B=0$.

The figure below draws exactly this stack: the dashed line is the mid-plane ($z=0$), the two teal blocks are the plies, and the two coloured arrows show each ply's contribution to $B$ — equal in size, opposite in sign, so they cancel to zero.

Figure Ex6 — symmetric [0/0] stack. Bottom ply contributes $-0.0625Q/2$ (orange arrow, below mid-plane), top ply contributes $+0.0625Q/2$ (plum arrow, above mid-plane). The two are mirror opposites, so the sum $B=0$.

1. Set the $z$ coordinate. Mid-plane at $z=0$. Total thickness $2t=0.5$ mm, so faces at $z=\pm 0.25$ mm. Ply 1 (bottom): $z_0=-0.25\to z_1=0$. Ply 2 (top): $z_1=0\to z_2=+0.25$ mm. Why this step? $B$ is measured from the mid-plane; the sign of $z$ is the whole point, so we must pin down where each ply sits relative to $z=0$.
2. $B$ formula (per stiffness $\bar{Q}$). $B_{ij}=\frac{1}{2}{\sum }_k{\bar{Q}}_{ij,k}(z_k^2-z_{k-1}^2)$. For a symmetric $[0/0]$ both plies share the same $\bar{Q}$ (same angle, from the definition above); call it $Q$. Why $z^2$? $B$ couples in-plane stretching to bending; the moment arm brings one factor of $z$ and the through-thickness integration of the linear strain brings another, so the integrand carries $z^2$.
3. Sum the two plies. $B=\frac{1}{2}Q[(0^2-(-0.25{)}^2)+(0.25^2-0^2)]=\frac{1}{2}Q[(-0.0625)+(0.0625)]=0.$ Why the cancellation? The bottom ply contributes $-0.0625Q\cdot \frac{1}{2}$ and the top ply $+0.0625Q\cdot \frac{1}{2}$ — exact mirror opposites, exactly the two arrows in the figure.
4. Contrast $[0/90]$. There $\bar{Q}$ differs between the two plies (a $90^{\circ }$ ply is stiff where the $0^{\circ }$ is soft, so its $\bar{Q}$ is different). The $\pm z$ terms no longer share the same $\bar{Q}$, so they don't cancel → $B\ne 0$ → the panel warps when loaded or heated. Why? The cancellation in step 3 relied on identical $\bar{Q}$ above and below; break that symmetry and the two arrows no longer match, leaving a leftover.

Verify: Units — $\bar{Q}$ [GPa] × mm² gives GPa·mm² for $B$, and $B=0$ is dimensionless-safe. The result matches the parent's rule "symmetric ⇒ $B=0$." ✓

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Ex 7 — Same ply, surface vs centre: bending weighting (cell C7)

Forecast: the parent note says $D\sim z^3$ so outer plies dominate. Guess: the surface position gives many times more bending stiffness — maybe 30–40×.

The figure shows the two placements side by side: the same ply (same $\bar{Q}$, same thickness) drawn once near the surface (orange) and once at the centre (teal), with each one's computed $D$ labelled. Notice only its height on the $z$ axis changed.

Figure Ex7 — identical ply, two positions. Surface placement gives $D=5.07$ GPa·mm³ (orange); central placement gives $D=0.18$ GPa·mm³ (teal). Same material, 28× stiffer in bending purely from the $z^3$ weighting.

1. $D$ formula (one ply). $D=\frac{1}{3}\bar{Q}(z_k^3-z_{k-1}^3)$, using the definition above with $\bar{Q}=139$ GPa. Why $z^3$? Bending stress grows with $z$ (outer fibres strain most, $=z\kappa$), its moment about the mid-plane adds another $z$, and integrating a $z^2$ integrand gives $z^3/3$.
2. Outer position. $D_{\text{out}}=\frac{1}{3}(139)[(0.5{)}^3-(0.25{)}^3]=\frac{1}{3}(139)(0.125-0.015625)=\frac{1}{3}(139)(0.109375)=5.068$ GPa·mm³. Why? We plug the two outer faces into the cube formula; the large $z$ values ($0.5$) make the bracket sizeable.
3. Central position. $D_{\text{cen}}=\frac{1}{3}(139)[(0.125{)}^3-(-0.125{)}^3]=\frac{1}{3}(139)(0.001953+0.001953)=\frac{1}{3}(139)(0.003906)=0.181$ GPa·mm³. Why the $-(-0.125{)}^3$? A central ply spans both signs of $z$; $(-0.125{)}^3=-0.001953$, and subtracting a negative adds it, but both cubes are tiny because $z$ is small near the middle.
4. Ratio. $D_{\text{out}}/D_{\text{cen}}=5.068/0.181=28.0$. Why? We divide the two to size the effect; the surface placement is 28× stiffer despite using the exact same material.

Verify: Same amount of the same material — only its distance from the mid-plane changed — yet it's 28× stiffer in bending on the surface. This is why aerospace skins put $0^{\circ }$ plies near the faces. ✓ Units GPa·mm³ throughout.

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Ex 8 — Real aircraft skin: choose the material per kilogram (cell C8)

Forecast: CFRP is both stiffer and lighter, so it should crush aluminium on stiffness-per-kg. Guess: roughly 3× better.

1. Define the figure of merit. Specific stiffness $=E/\rho$ — stiffness delivered per unit density. Aerospace maximises this because every kilogram costs fuel. See Specific strength and stiffness in aerospace materials. Why this tool, not just $E$? Two panels of equal $E$ but different $\rho$ are not equal — the lighter one lets you fly farther. $E/\rho$ folds both in.
2. CFRP. $E/\rho =139/1.6=86.9$ (GPa per g/cm³). Why? We divide stiffness by density; CFRP's low density (1.6) boosts the ratio.
3. Aluminium. $E/\rho =70/2.7=25.9$. Why? Same division for the metal; its higher density (2.7) and lower $E$ both drag the ratio down.
4. Ratio. $86.9/25.9=3.35$. Why? We compare the two figures of merit directly; CFRP delivers 3.35× the stiffness for the same weight.

Verify: CFRP gives 3.35× the stiffness per kilogram. This is the whole reason the Boeing 787 is ~50% composite by weight. Caveat from the parent: this is the longitudinal value — you must lay up at angles (quasi-isotropic) to keep this advantage in every direction, which lowers the effective number but still beats aluminium. ✓

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Ex 9 — Exam twist: the wrong-formula trap + over-limit $V_f$ (cell C9)

Forecast: transverse ⇒ series ⇒ must use the inverse rule (error 1). And $V_f=0.75$ is above the practical packing limit (~0.65), so the ply couldn't be made properly (error 2). Guess: correct $E_2$ is single-digit GPa.

1. Error 1 — wrong physics. Transverse loading is iso-stress (series), so the parallel rule $E_f{V}_f+E_m{V}_m$ (iso-strain) does not apply. Ask the parent note's question: "parallel or series for this load?" → series. Why it matters? The parallel rule is an upper bound; using it transversely overstates stiffness enormously ($173$ vs the truth below).
2. Error 2 — impossible $V_f$. Above $V_f\approx 0.65$–$0.70$ there isn't enough matrix to wet and bond all fibres → voids and poor load transfer. $V_f=0.75$ is not manufacturable. Why? Fibres are round; they can't pack to 100% and still leave glue between them. See Fibre–matrix interface and load transfer.
3. Correct calculation (using a feasible $V_f=0.65$, $V_m=0.35$): $\frac{1}{E_2}=\frac{0.65}{230}+\frac{0.35}{3}=0.002826+0.116667=0.119493\Rightarrow E_2=8.37\text{ GPa}.$ Why? We now use the correct series rule with a manufacturable fibre fraction; adding the two compliances and inverting lands near the soft matrix value.
4. Contrast the two answers. The student's $173$ GPa is off by a factor of ~21 from the physically correct $\sim 8$ GPa. Why? We divide $173/8.37\approx 21$ to show the size of the blunder — an entire order of magnitude, not a rounding slip.

Verify: $E_2$ still lies between $E_m=3$ and $E_f=230$ (any mixture must) and near the soft end — correct for iso-stress. The trap is settled: right formula and feasible $V_f$. ✓

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Recall Self-test: name the cell, then solve

Load runs along the fibres — which formula and why? ::: Iso-strain (parallel), $E_1=E_f{V}_f+E_m{V}_m$; bonded phases stretch equally. Load runs across the fibres — which formula and why? ::: Iso-stress (series), $1/E_2=V_f/E_f+V_m/E_m$; load passes through the soft matrix. At $V_f=0$ what does either formula give? ::: $E_m$ — pure matrix, no anisotropy. Why does $B=0$ for a symmetric lay-up? ::: Each $+z$ ply has an identical $-z$ twin; the $z^2$ terms cancel exactly. Why do outer plies dominate bending? ::: $D\sim z^3$, so distance from the mid-plane counts cubically; surface plies strain most under bending. Best single figure of merit for an aircraft skin? ::: Specific stiffness $E/\rho$ — stiffness per unit weight.

Connections

• 5.4.06 Polymer-matrix composites — CFRP, GFRP; ply lay-up, laminate theory (Hinglish)
• Stress, strain and Young's modulus
• Anisotropy and crystal directions
• Fibre–matrix interface and load transfer
• Specific strength and stiffness in aerospace materials
• Fracture and fatigue in composites
• Thermoset vs thermoplastic polymers