Polymer-matrix composites — CFRP, GFRP; ply lay-up, laminate theory: Visual walkthrough

Step 1 — What a single ply is, and the two directions

WHAT. A ply is a thin sheet of many parallel fibres, all pointing one way, drowned in soft glue (the matrix). Two directions matter and they behave completely differently:

Direction 1 = along the fibres (the "1-direction").
Direction 2 = across the fibres, in the plane of the sheet (the "2-direction").

WHY. Because the fibres are the only stiff thing here, which way you pull relative to them decides everything. Pulling along fibre bonds is like pulling a taut rope; pulling across means pulling on the jelly between ropes. Same material — wildly different answers. That direction-dependence is exactly anisotropy.

PICTURE. Below: the ply, its fibres (coral rods) in matrix (lavender). The green arrow is the 1-direction (along), the butter arrow is the 2-direction (across).

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

Step 2 — Pull ALONG the fibres: they must stretch together

WHAT. Grab the ply at both ends and pull along direction 1. The fibres and matrix are glued, so their ends move together — they have to stretch by the same fraction.

WHY. If the fibre stretched more than the matrix beside it, the glue joint would have to tear. It doesn't (good interface — see Fibre–matrix interface and load transfer). So the strain is shared and equal. We name this the iso-strain condition:

$ε_{1} = ε_{f} = ε_{m}$

$ε_{1}$ = the whole ply's stretch,
$ε_{f}$ = the fibre's stretch,
$ε_{m}$ = the matrix's stretch,
the "=" signs say all three are the same number.

PICTURE. Two bars side by side (fibre + matrix), clamped between the same two grips. When the right grip moves, both bars lengthen by the identical amount — that shared elongation is iso-strain.

Step 3 — Force adds up across the cross-section

WHAT. The total pull on the ply is just the fibre's share of pull plus the matrix's share:

$F = F_{f} + F_{m}$

Rewrite each force as stress × area (because $σ = F / A$ , so $F = σ A$ ):

$σ_{1} A = σ_{f} A_{f} + σ_{m} A_{m}$

$σ_{1} A$ = total force = ply stress × ply area,
$σ_{f} A_{f}$ = force carried by fibres = fibre stress × fibre area,
$σ_{m} A_{m}$ = force carried by matrix.

WHY. Forces on parallel load paths simply add — nothing is lost. The cross-section is split between fibre area $A_{f}$ and matrix area $A_{m}$ ; each patch of area pushes back with its own stress.

PICTURE. The end-on view of the ply: circles = fibre cross-sections, lavender fill = matrix. The total area $A$ is chopped into $A_{f}$ (all circles) and $A_{m}$ (everything else). Because fibres run the full length, area fraction = volume fraction: $A_{f} / A = V_{f}$ .

Divide the force equation by the total area $A$ , and use $A_{f} / A = V_{f}$ , $A_{m} / A = V_{m}$ :

$σ_{1} = σ_{f} V_{f} + σ_{m} V_{m}$

this says: the ply's stress is a volume-weighted average of the two phase stresses.

Step 4 — Turn stress into stiffness: the Rule of Mixtures

WHAT. Each phase obeys Hooke's law $σ = E ε$ . Substitute for every stress in the Step-3 result:

$σ_{1} E_{1} ε_{1} = σ_{f} E_{f} ε_{f} V_{f} + σ_{m} E_{m} ε_{m} V_{m}$

WHY. We want stiffness $E_{1}$ , not stress. Hooke's law is the bridge from "how hard I push" ( $σ$ ) to "how stiff it is" ( $E$ ). Now use the Step-2 gift: all strains are equal, $ε_{1} = ε_{f} = ε_{m}$ , so we can divide the whole line by that common $ε$ and it vanishes:

$E_{1} = E_{f} V_{f} + E_{m} V_{m}$

$E_{1}$ = ply stiffness along fibres,
$E_{f} V_{f}$ = fibre's contribution, weighted by how much fibre there is,
$E_{m} V_{m}$ = matrix's contribution.

This is the Rule of Mixtures. It is a plain weighted average → an upper bound on stiffness. Because $E_{f} ≫ E_{m}$ , the $E_{f} V_{f}$ term dominates: the fibres carry the load.

PICTURE. A bar chart: fibre bar (tall, $E_{f}$ ) scaled by $V_{f}$ , matrix bar (tiny, $E_{m}$ ) scaled by $V_{m}$ , stacked to give $E_{1}$ . You can literally see the fibres owning the stiffness.

Step 5 — Now pull ACROSS the fibres: same force, different stretch

WHAT. Rotate the pull by 90° (direction 2). Now the load must pass through the fibre and through the matrix in a chain — they are in series. In a series chain every link feels the same force, hence the same stress (equal cross-sections):

$σ_{2} = σ_{f} = σ_{m} (iso-stress)$

WHY. Picture the load line as an arrow entering the matrix, crossing into a fibre, back into matrix — one after another. Whatever force enters one link exits into the next unchanged (no force disappears). So stress is shared equally, but stretch is not: the soft matrix stretches a lot, the stiff fibre barely at all.

PICTURE. Two bars stacked end to end (in series) between the grips. Same tension pulls both, but the soft (lavender) block visibly elongates far more than the stiff (coral) block.

Step 6 — Strains add up along the chain → the inverse rule

WHAT. In series the stretches add. Over a unit cell of the ply, the total across-fibre strain is the volume-weighted sum of each phase's strain:

$ε_{2} = V_{f} ε_{f} + V_{m} ε_{m}$

$ε_{2}$ = whole ply's across-strain,
$V_{f} ε_{f}$ = fibre's stretch weighted by how much thickness is fibre,
$V_{m} ε_{m}$ = matrix's stretch weighted by its thickness share.

WHY. Displacements in a chain accumulate — walk through the fibre part, then the matrix part; the total distance is the sum. Now swap every strain for $ε = σ / E$ (Hooke's law rearranged), and cancel the common stress $σ$ (from Step 5, $σ_{2} = σ_{f} = σ_{m}$ ):

$\frac{σ}{E _{2}} = V_{f} \frac{σ}{E _{f}} + V_{m} \frac{σ}{E _{m}} ⟹ \frac{1}{E _{2}} = \frac{V _{f}}{E _{f}} + \frac{V _{m}}{E _{m}}$

the reciprocals appear because in series it is compliances ( $1/ E$ , the "easiness to stretch") that add, not stiffnesses.

This is a harmonic average → a lower bound. A soft link in series drags the whole answer down, so $E_{2} ≪ E_{1}$ .

PICTURE. A "how easy to stretch" bar (compliance $1/ E$ ): matrix compliance is huge, fibre compliance tiny; adding them (weighted) is dominated by the matrix → the resulting $E_{2}$ is small.

Step 7 — Put numbers in and watch the anisotropy

WHAT. Carbon/epoxy ply: $E_{f} = 230$ GPa, $E_{m} = 3$ GPa, $V_{f} = 0.6$ (so $V_{m} = 0.4$ ).

Along the fibres (Step 4): $E_{1} = 230 (0.6) + 3 (0.4) = 138 + 1.2 = 139.2 GPa$

Across the fibres (Step 6): $E_{2} = (\frac{0.6}{230} + \frac{0.4}{3})^{- 1} = (0.002609 + 0.133333)^{- 1} = 7.36 GPa$

WHY. These two lines are the entire justification for stacking plies at angles. The ratio

$\frac{E _{1}}{E _{2}} = \frac{139.2}{7.36} \approx 18.9$

means the ply is nearly 19× stiffer one way than the other. One ply is useless off-axis — see Specific strength and stiffness in aerospace materials and Fracture and fatigue in composites for why that matters.

PICTURE. A polar "stiffness rose": long lobe along $0°$ (the $E_{1}$ direction), squashed almost flat near $90°$ (the $E_{2}$ direction). The shape is the anisotropy.

Step 8 — The edge cases (so no scenario surprises you)

WHAT & WHY — every limit checked:

Worked example Degenerate and limiting inputs

$V_{f} = 0$ (all matrix): $E_{1} = E_{m} = 3$ , and $1/ E_{2} = 1/ E_{m} \Rightarrow E_{2} = E_{m} = 3$ . Both rules collapse to plain resin. ✔ (A block of glue is the same stiffness whichever way you pull — isotropic.)
$V_{f} = 1$ (all fibre): $E_{1} = E_{f} = 230$ , $E_{2} = E_{f} = 230$ . Both give the bare fibre. ✔
$E_{f} = E_{m}$ (fibre no stiffer than glue): $E_{1} = E_{m} (V_{f} + V_{m}) = E_{m}$ and $E_{2} = E_{m}$ . If the phases are identical, direction can't matter — anisotropy vanishes. ✔
$E_{m} \to 0$ (glue infinitely soft): $E_{1} \to E_{f} V_{f}$ (fibres still carry along-load), but $1/ E_{2} \to \infty$ , so $E_{2} \to 0$ . Across the fibres, an infinitely soft matrix means zero transverse stiffness — nothing holds it. This is the extreme that proves the mistake "use $E_{1}$ for both directions" catastrophically wrong.
Bounds sandwich: for any $0 < V_{f} < 1$ with $E_{f} > E_{m}$ , the true stiffness in a mixed load lies between $E_{2}$ (lower, series) and $E_{1}$ (upper, parallel). Reality never escapes this envelope.

PICTURE. $E_{1}$ (upper curve) and $E_{2}$ (lower curve) plotted against $V_{f}$ from 0 to 1; they pin together at both ends ( $V_{f} = 0$ and $V_{f} = 1$ ) and bulge apart in the middle — the shaded gap is the achievable anisotropy.

The one-picture summary

Parallel vs series is the whole story: same strain → stiffnesses add (Rule of Mixtures, big $E_{1}$ ); same stress → compliances add (inverse rule, small $E_{2}$ ). One picture holds both, side by side.

Recall Feynman: the whole walkthrough in plain words

Take a bundle of stiff straws glued into jelly. Pull along the straws: the glue forces every straw to stretch the same tiny amount, so all the straws pull hard together — you feel almost the full straw stiffness. That "everyone stretches the same, forces add" is parallel, and it gives a big stiffness that's basically a weighted average, $E_{1} = E_{f} V_{f} + E_{m} V_{m}$ . Now pull sideways, across the straws: the pull has to squeeze through jelly, then straw, then jelly, like beads on a string. Every bead feels the same tug, but the soft jelly stretches like crazy while the straw hardly moves. Because the stretches add up along the chain and the softest link wins, the sideways stiffness is small — you add the "easinesses" ( $1/ E$ ) instead of the stiffnesses. That's the inverse rule, $1/ E_{2} = V_{f} / E_{f} + V_{m} / E_{m}$ . Plug in real carbon/epoxy and you get 139 vs 7 GPa — nearly 19 times stiffer one way than the other. One ply is a one-trick pony. So engineers stack plies turned to different angles, and that stack is strong every way. Everything in laminate theory grows from these two little pictures: parallel adds stiffness, series adds softness.

Connections

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