Composite materials — fiber-matrix, ply properties, laminate theory: Exercises

This is your self-testing sheet for the parent topic on composites (fiber–matrix, ply properties, laminate theory). Work each problem before opening the solution. Problems climb from "can you recall the formula" to "can you design a layup and defend it".

Throughout, we reuse these symbols exactly as the parent note defined them. A quick reminder so no symbol is used before you can picture it:

Recall Symbol dictionary (open if any symbol looks unfamiliar)

$E_{f}$ ::: Young's modulus (stiffness) of the fiber alone — how hard it resists stretching, in GPa. $E_{m}$ ::: Young's modulus of the matrix (the glue/resin) alone. $V_{f}, V_{m}$ ::: Volume fractions: what slice of the material is fiber vs matrix. They add to 1. $E_{1}$ ::: Composite stiffness measured along the fibers (the strong direction). $E_{2}$ ::: Composite stiffness measured across the fibers (the weak direction). $G_{12}$ ::: In-plane shear modulus — resistance to the material "scissoring". $ν_{12}$ ::: Major Poisson's ratio — pull along fibers, how much it thins sideways. $[Q]$ ::: Ply stiffness matrix in material axes (1 = fiber, 2 = transverse). $[\overset{ˉ}{Q}]$ ::: The SAME ply's stiffness rotated to the laminate x–y axes when fibers sit at angle $θ$ . $θ$ ::: Angle between the fiber direction and the laminate x-axis.

Level 1 — Recognition

L1.1 — Name the rule

Problem. You pull a unidirectional composite along its fibers. Which averaging rule gives $E_{1}$ , and why is it a direct (not inverse) sum?

Recall Solution

Rule of Mixtures: $E_{1} = E_{f} V_{f} + E_{m} V_{m}$ . Look at the fibers and matrix as two springs standing side by side (parallel) — pull the whole cube and both stretch the same amount (equal strain). Parallel springs add their stiffnesses directly, so stiffness contributions add in proportion to how much of each phase is present. Hence a direct weighted sum.

L1.2 — Match the direction to the formula

Problem. State which formula (direct or inverse Rule of Mixtures) applies to each: (a) $E_{1}$ , (b) $E_{2}$ , (c) $G_{12}$ .

Recall Solution

(a) $E_{1}$ → direct (parallel springs, equal strain). (b) $E_{2}$ → inverse: $\frac{1}{E _{2}} = \frac{V _{f}}{E _{f}} + \frac{V _{m}}{E _{m}}$ (series springs, equal stress). (c) $G_{12}$ → inverse: $\frac{1}{G _{12}} = \frac{V _{f}}{G _{f}} + \frac{V _{m}}{G _{m}}$ (also series-like: load must pass through the soft matrix).

Figure — Composite materials — fiber-matrix, ply properties, laminate theory

L1.3 — Read off the anisotropy

Problem. A ply has $E_{1} = 139.2$ GPa and $E_{2} = 7.4$ GPa. Compute the anisotropy ratio $E_{1} / E_{2}$ and say in one sentence what it means physically.

Recall Solution

$E_{1} / E_{2} = 139.2/7.4 \approx 18.8$ . The material is about 19× stiffer along the fibers than across them — it is strongly anisotropic, so orientation matters enormously.

Level 2 — Application

L2.1 — Longitudinal modulus

Problem. Glass–epoxy: $E_{f} = 72$ GPa, $E_{m} = 3.5$ GPa, $V_{f} = 0.55$ . Find $E_{1}$ .

Recall Solution

$V_{m} = 1 - 0.55 = 0.45$ . $E_{1} = 72 (0.55) + 3.5 (0.45) = 39.6 + 1.575 = 41.175 GPa .$ So $E_{1} \approx 41.2$ GPa.

L2.2 — Transverse modulus

Problem. Same glass–epoxy ( $E_{f} = 72$ , $E_{m} = 3.5$ , $V_{f} = 0.55$ ). Find $E_{2}$ .

Recall Solution

$\frac{1}{E _{2}} = \frac{0.55}{72} + \frac{0.45}{3.5} = 0.007639 + 0.128571 = 0.136210.$ $E_{2} = \frac{1}{0.136210} = 7.342 GPa .$ Notice: even with 55% stiff fiber, the transverse stiffness collapses toward the soft matrix, because the load is forced through the weak resin in series.

L2.3 — Minor Poisson's ratio

Problem. A ply has $ν_{12} = 0.28$ , $E_{1} = 130$ GPa, $E_{2} = 9$ GPa. Find $ν_{21}$ .

Recall Solution

Symmetry of the compliance matrix demands $ν_{21} = ν_{12} E_{2} / E_{1}$ . $ν_{21} = 0.28 \times \frac{9}{130} = 0.28 \times 0.069231 = 0.019385 \approx 0.0194.$ The minor ratio is tiny — pulling across the fibers barely contracts the stiff fiber direction.

Level 3 — Analysis

L3.1 — Build the ply stiffness matrix $[Q]$

Problem. For a ply with $E_{1} = 140$ GPa, $E_{2} = 10$ GPa, $G_{12} = 5$ GPa, $ν_{12} = 0.30$ , compute $Q_{11}, Q_{22}, Q_{12}, Q_{66}$ .

Recall Solution

Step 1 — minor Poisson's ratio. $ν_{21} = 0.30 \times \frac{10}{140} = 0.021429.$ Step 2 — the coupling denominator. (This term appears in every $Q$ ; it accounts for the fact that stretching one way induces stress the other way, so stiffness is slightly boosted.) $1 - ν_{12} ν_{21} = 1 - 0.30 (0.021429) = 1 - 0.006429 = 0.993571.$ Step 3 — the entries. $Q_{11} = \frac{140}{0.993571} = 140.906 GPa, Q_{22} = \frac{10}{0.993571} = 10.065 GPa,$ $Q_{12} = \frac{0.30 \times 10}{0.993571} = 3.0194 GPa, Q_{66} = G_{12} = 5 GPa .$ $[Q] = 140.9 3.02 0 3.02 10.06 0 005 GPa .$ $Q_{11}$ is barely above $E_{1}$ — the Poisson boost is only 0.6%. But $Q_{12} \neq = 0$ tells you pulling in direction 1 also generates stress in direction 2.

L3.2 — Degenerate case: what happens at $θ = 0°$ ?

Problem. Show that the transformation matrix $[T]$ reduces to the identity when $θ = 0°$ , and explain what that means for $[\overset{ˉ}{Q}]$ .

Recall Solution

At $θ = 0$ : $cos θ = 1$ , $sin θ = 0$ , so $cos^{2} θ = 1$ , $sin^{2} θ = 0$ , $sin θ cos θ = 0$ . $[T] = 100010001 = [I] .$ Then $[\overset{ˉ}{Q}] = [T]^{- 1} [Q] [T]^{- T} = [I] [Q] [I] = [Q]$ . Meaning: when the fibers are already aligned with the laminate x-axis, there is nothing to rotate — the material axes and laminate axes coincide, so $[\overset{ˉ}{Q}] = [Q]$ . This is the sanity check every transformation must pass.

L3.3 — Off-axis coupling at $θ = 45°$

Problem. Without computing the full matrix, explain which new coupling term appears in $[\overset{ˉ}{Q}]$ at $θ = 45°$ that was zero in $[Q]$ , and what physical behaviour it causes. Then compute $sin θ cos θ$ and $cos^{2} θ - sin^{2} θ$ at $45°$ to justify.

Recall Solution

At $45°$ : $sin θ cos θ = \frac{1}{2}$ (its maximum) and $cos^{2} θ - sin^{2} θ = 0$ . The terms $\overset{ˉ}{Q}_{16}$ and $\overset{ˉ}{Q}_{26}$ — the extension–shear coupling terms — become non-zero (they were exactly 0 in the material-axis $[Q]$ ). Physical meaning: pull the laminate straight along x, and it shears (the square distorts into a parallelogram). This is why a single off-axis ply "twists" under simple tension — a behaviour metals never show.

Level 4 — Synthesis

L4.1 — Specific stiffness trade study

Problem. Compare specific stiffness $E / ρ$ (units GPa per g/cm³) for:

Carbon–epoxy: $E_{1} = 139.2$ GPa, $ρ = 1.60$ g/cm³
Aluminium 7075: $E = 71.7$ GPa, $ρ = 2.81$ g/cm³ Which wins, and by what factor? Then state one direction where the composite loses.

Recall Solution

Composite: $139.2/1.60 = 87.0$ GPa·cm³/g. Aluminium: $71.7/2.81 = 25.5$ GPa·cm³/g. Ratio: $87.0/25.5 = 3.41$ → the composite is ~3.4× more efficient along its fibers. Where it loses: across the fibers, $E_{2} \approx 7.4$ GPa, giving specific stiffness $7.4/1.60 = 4.6$ — far below aluminium's isotropic 25.5. So a naive single-direction layup is worse than aluminium in the transverse direction. This is the reason for multi-directional laminates and sandwich cores rather than raw unidirectional plies. Feeds directly into mass budgeting.

L4.2 — Designing a quasi-isotropic layup

Problem. You must build a panel that is (approximately) equally stiff in all in-plane directions, using the ply from L3.1. State a standard ply-angle set that achieves this and explain why those angles.

Recall Solution

A quasi-isotropic stack: $[0°/ + 45° / - 45° /90°]_{s}$ (the $_{s}$ means mirror it for symmetry). Why these angles: to average stiffness evenly around the 180° of in-plane directions, you place equal numbers of plies at evenly spaced angles. Four angles spaced 45° apart ( $0, 45, 90, 135 \equiv - 45$ ) tile the half-circle uniformly, so the averaged in-plane modulus becomes (nearly) direction-independent. The $\pm 45$ pair is balanced, cancelling the extension–shear coupling from L3.3; the mirror symmetry ( $_{s}$ ) cancels bending–stretching coupling. Verified in practice with finite-element analysis.

Level 5 — Mastery

L5.1 — Back-solve the fiber volume fraction

Problem. A carbon–epoxy ply is measured to have $E_{1} = 120$ GPa. With $E_{f} = 230$ GPa and $E_{m} = 3$ GPa, what fiber volume fraction $V_{f}$ was achieved in manufacturing?

Recall Solution

Start from Rule of Mixtures with $V_{m} = 1 - V_{f}$ : $E_{1} = E_{f} V_{f} + E_{m} (1 - V_{f}) = E_{m} + (E_{f} - E_{m}) V_{f} .$ Solve for $V_{f}$ : $V_{f} = \frac{E _{1} - E _{m}}{E _{f} - E _{m}} = \frac{120 - 3}{230 - 3} = \frac{117}{227} = 0.5154.$ So $V_{f} \approx 0.52$ — a slightly low fraction, hinting at under-consolidation or excess resin, worth flagging to material selection / QA.

L5.2 — Full limiting-behaviour sweep of $E_{2}$

Problem. For the transverse modulus $\frac{1}{E _{2}} = \frac{V _{f}}{E _{f}} + \frac{V _{m}}{E _{m}}$ with $E_{f} = 230$ , $E_{m} = 3$ GPa, evaluate the two limiting cases $V_{f} \to 0$ and $V_{f} \to 1$ , and explain why $E_{2}$ is dominated by the matrix over most of the practical range.

Recall Solution

Case $V_{f} \to 0$ (all matrix): $\frac{1}{E _{2}} = \frac{0}{230} + \frac{1}{3} \Rightarrow E_{2} = 3$ GPa. Pure matrix — correct. Case $V_{f} \to 1$ (all fiber): $\frac{1}{E _{2}} = \frac{1}{230} + 0 \Rightarrow E_{2} = 230$ GPa. Pure fiber — correct. Why the matrix dominates in between: at a typical $V_{f} = 0.6$ , $\frac{1}{E _{2}} = \frac{0.6}{230} + \frac{0.4}{3} = 0.002609 + 0.133333 = 0.135942, E_{2} = 7.36 GPa .$ The fiber term ( $0.0026$ ) is ~50× smaller than the matrix term ( $0.133$ ). In a series path, the softest element controls compliance (like the widest gap in a chain of springs), so $E_{2}$ hugs the matrix value until $V_{f}$ is nearly 1.

L5.3 — Consistency check via strain energy

Problem. Confirm that the two expressions for $Q_{12}$ , namely $\frac{ν _{12} E _{2}}{1 - ν _{12} ν _{21}}$ and $\frac{ν _{21} E _{1}}{1 - ν _{12} ν _{21}}$ , are equal — using $E_{1} = 140$ , $E_{2} = 10$ GPa, $ν_{12} = 0.30$ . Explain why they must be equal.

Recall Solution

$ν_{21} = 0.30 \times 10/140 = 0.0214286$ ; denominator $= 0.993571$ (from L3.1). First form: $\frac{0.30 \times 10}{0.993571} = \frac{3.0}{0.993571} = 3.01942$ GPa. Second form: $\frac{0.0214286 \times 140}{0.993571} = \frac{3.0}{0.993571} = 3.01942$ GPa. They match. Why they must: the reciprocity relation $ν_{12} E_{2} = ν_{21} E_{1}$ (from L2.3) makes the two numerators identical. This is not a coincidence — it enforces that the ply's stiffness matrix is symmetric, which is required for the material to store strain energy consistently (no free energy from a loading cycle).

Self-test cloze recap

The direct Rule of Mixtures applies because along the fibers, fiber and matrix act as parallel springs at equal strain. The transverse modulus $E_{2}$ uses the inverse Rule of Mixtures because the phases act in series at equal stress. At $θ = 45°$ the new non-zero coupling terms $\overset{ˉ}{Q}_{16}, \overset{ˉ}{Q}_{26}$ cause the panel to extend and shear at the same time. Reciprocity $ν_{12} E_{2} = ν_{21} E_{1}$ guarantees the stiffness matrix is symmetric.