Classical laminate theory — ABD matrix: Visual walkthrough

Step 0 — The one picture everything sits on

WHAT. Before any equation, fix the ruler. Our laminate is a stack of thin flat layers glued together, and we call the total number of layers $n$ (so the plies are numbered $k = 1, 2, \dots, n$ from bottom to top). We draw a vertical arrow — call it $z$ — pointing up through the stack. Its zero is placed exactly at the geometric middle, the mid-plane.

WHY here. Every quantity below is measured relative to that mid-plane: a layer's height, how far a fibre sits above or below centre, the lever arm of a force. Choosing $z = 0$ at the middle is not decoration — it is the choice that makes symmetry cause things to cancel later. Get this wrong and the whole $[B]$ story breaks.

PICTURE. Look at the figure. The total thickness is $h$ , so the top face is at $z = + h /2$ and the bottom at $z = - h /2$ . Each layer $k$ spans from a lower edge $h_{k - 1}$ to an upper edge $h_{k}$ . The four coloured bands are four plies (so here $n = 4$ ); the dashed white line is the mid-plane where $z = 0$ .

Figure — Classical laminate theory — ABD matrix

Step 1 — Freeze the geometry: "plane sections stay plane"

WHAT. Draw a straight vertical line through the unloaded plate — a column of material. Now bend and stretch the plate. The kinematic assumption of classical laminate theory says that line stays straight and stays perpendicular to the mid-surface; it may tilt and slide, but it never curls or warps.

WHY this and not full 3D. A real solid could deform in wildly complicated ways through its thickness. Solving that needs 3D elasticity in every ply — hopeless by hand and expensive on a computer. The "plane stays plane" rule (the Kirchhoff–Love assumption) is a deliberate simplification that is accurate while the plate is thin (thickness ≪ width). It collapses the through-thickness behaviour into just two ingredients: how the middle moves, and how the plate tilts.

PICTURE. The figure shows the same vertical fibre before (grey) and after loading (blue). Two things happened: the middle slid sideways by a uniform amount — that is membrane strain — and the line tilted, more tilt meaning more curvature. The total strain at height $z$ is the sum of a constant part plus a part that grows straight-line with $z$ .

Formula Strain varies straight-line through the thickness

$strain at height z ϵ (z) = strain at the middle ϵ^{0} + z tilt per unit height κ$

$ϵ^{0} = [ϵ_{x}^{0}, ϵ_{y}^{0}, γ_{x y}^{0}]^{T}$ ::: the three mid-plane strains — the constant offset, same at every height. Here $ϵ_{x}^{0}$ and $ϵ_{y}^{0}$ are direct stretches (fractional length change) along $x$ and $y$ , while $γ_{x y}^{0}$ is the engineering shear strain — the total change of the right angle between the $x$ and $y$ fibres, in radians (twice the tensor shear).
$κ = [κ_{x}, κ_{y}, κ_{x y}]^{T}$ ::: the three curvatures, units $1/ m$ ; multiplied by $z$ they say "the higher you are, the more the bending adds or subtracts strain." Here $κ_{x}$ and $κ_{y}$ are bending curvatures (how sharply the plate cups about the $y$ and $x$ axes), while $κ_{x y}$ is the twist curvature — how much the plate warps like a propeller, one corner up and the diagonal corner down.
$z$ ::: the lever that turns curvature into extra strain. At $z = 0$ (mid-plane) it vanishes, so $ϵ (0) = ϵ^{0}$ — exactly why we put zero at the middle.

The equation is a straight line in $z$ : intercept $ϵ^{0}$ , slope $κ$ .

Step 2 — Each ply answers with its own stress

WHAT. Strain is a shape change; stress is the push-back force per area. Inside layer $k$ these are linked by that layer's stiffness matrix $[Q]_{k}$ : multiply the strain and you get the stress.

WHY $[Q]$ carries a subscript $k$ . Two plies can be cut from the very same carbon/epoxy and still push back differently, because their fibres point different ways. A ply with fibres along $x$ is stiff along $x$ ; rotate the ply $90°$ and it is now stiff along $y$ . That rotation is baked into $[Q]_{k}$ via the fibre angle $θ_{k}$ — see Reduced Stiffness Matrix Q for where $[Q]$ itself comes from and Transformation of Stiffness for how the angle rotates it.

PICTURE. The strain profile from Step 1 is one smooth straight line through the stack (left). Feeding it ply-by-ply through different $[Q]_{k}$ gives a stress profile that jumps at every glue line (right): same strain, but a stiffer ply reacts with more stress. Those jumps are the whole reason a laminate is interesting.

Step 3 — Add up the stresses to get force: the $[A]$ and $[B]$ matrices

WHAT. We cannot measure stress at every height in practice; we can measure the total in-plane force per unit width $N$ . Its three components are $N = [N_{x}, N_{y}, N_{x y}]^{T}$ — $N_{x}$ and $N_{y}$ are the direct pulls along $x$ and $y$ , and $N_{x y}$ is the in-plane shear force, each measured per metre of edge width (units N/m, i.e. force divided by the width the edge runs along). To get $N$ , add up (integrate) the stress over the whole thickness. Because $[Q]$ jumps between plies, the smooth integral becomes a sum of ply integrals.

WHY integrate — why this tool. A force is a stress spread over an area. "Spread over the thickness" is precisely what an integral $\int σ d z$ computes: it accumulates every thin slice's contribution. No cheaper tool exists — a plain sum would ignore that stress changes within a ply, and it changes because $z κ$ tilts it.

PICTURE. The figure shades the area under the stress-vs- $z$ curve. That shaded area is $N$ . Split it into the flat piece (driven by $ϵ^{0}$ ) and the sloped piece (driven by $κ$ ): the flat piece needs $\int d z = t_{k}$ , the sloped piece needs $\int z d z$ .

N = \int_{- h /2}^{h /2} σ d z = k = 1 \sum n [Q]_{k} = t_{k} \int_{h_{k - 1}}^{h_{k}} d z ϵ^{0} + k = 1 \sum n [Q]_{k} = \frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2}) \int_{h_{k - 1}}^{h_{k}} z d z κ

where the sum runs over all $n$ plies.

Why $[B]$ can vanish. The weight $\frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2})$ is negative for a ply below the mid-plane and positive for its mirror ply above. If the stack is a mirror image about $z = 0$ (a symmetric laminate), each pair cancels and $[B] = 0$ . See Symmetric and Balanced Laminates.

Step 4 — Add up the torques to get moment: the $[D]$ matrix

WHAT. A stress far from the mid-plane not only pushes, it twists the plate — it has a lever arm $z$ . The total bending moment per unit width $M = [M_{x}, M_{y}, M_{x y}]^{T}$ (units N, i.e. moment per metre of width) is the stress times its lever arm $z$ , integrated through the thickness.

WHY multiply by $z$ . Torque = force × distance from the pivot. Here the pivot is the mid-plane and the distance is $z$ . So every stress slice contributes $z σ d z$ to the moment. This one extra factor of $z$ is the entire difference between force and moment — and it is what makes outer plies dominate bending.

PICTURE. The figure weights the same stress profile by $z$ : near the mid-plane $z \approx 0$ so contributions are tiny; at the outer faces $z$ is largest so those plies dominate. The signed shaded area is $M$ . The strain term $z κ$ multiplied again by $z$ gives $z^{2}$ — always positive — hence $[D]$ is always positive-definite.

M = \int_{- h /2}^{h /2} z σ d z = k = 1 \sum n [Q]_{k} \frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2}) \int_{h_{k - 1}}^{h_{k}} z d z ϵ^{0} + k = 1 \sum n [Q]_{k} \frac{1}{3} (h_{k}^{3} - h_{k - 1}^{3}) \int_{h_{k - 1}}^{h_{k}} z^{2} d z κ

Step 5 — Snap the four blocks together

WHAT. Stack the force equation on top of the moment equation. Two $3$ -vectors of load, two $3$ -vectors of response, four $3 \times 3$ blocks — a single $6 \times 6$ machine.

WHY assemble. Individually $[A], [B], [D]$ answer sub-questions. Together they answer the real one: given any combination of forces and moments, what does the plate do? And the block layout makes the physics visible at a glance — the off-diagonal $[B]$ blocks are exactly the stretch–bend coupling.

PICTURE. The figure colours the $6 \times 6$ : blue $[A]$ top-left, green $[D]$ bottom-right, red $[B]$ on both off-diagonals. Load vector on the right, response on the far right.

Units check (each block earns its power of length by how many $z$ 's it integrated): $[A]$ is $Pa\cdotm$ , $[B]$ is $Pa\cdotm^{2}$ , $[D]$ is $Pa\cdotm^{3}$ .

Step 6 — The edge cases you must never trip over

WHAT / WHY / PICTURE — all three cases in one figure. Real derivations die on the corner cases. Here are the three that always appear.

Symmetric stack ( $[B] = 0$ ). Mirror-image plies about $z = 0$ make every $\frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2})$ pair cancel. The $6 \times 6$ becomes block-diagonal: $ϵ^{0} = [A]^{- 1} N$ and $κ = [D]^{- 1} M$ decouple. Pull it, it just stretches; no surprise bending. This is why aerospace laminates are usually symmetric.
Antisymmetric $[0/90]$ ( $[B] \neq = 0$ ). Take two plies each $0.5 mm$ thick, the stiff $0°$ ply below centre (from $z = - 0.5 mm$ to $0$ ) and the compliant $90°$ ply above ( $0$ to $+ 0.5 mm$ ). The coupling term is $B_{11} = \frac{1}{2} [Q_{11}^{(0°)} (0^{2} - (- 0.5 \times 1 0^{- 3})^{2}) + Q_{11}^{(90°)} ((0.5 \times 1 0^{- 3})^{2} - 0^{2})] .$ Each squared height is in metres, so $(0.5 \times 1 0^{- 3} m)^{2} = 0.25 \times 1 0^{- 6} m^{2}$ — that is where the tiny factor $1 0^{- 6}$ comes from: it is simply $mm^{2}$ written in $m^{2}$ . The bottom ply contributes $- 0.25 \times 1 0^{- 6}$ and the top ply $+ 0.25 \times 1 0^{- 6}$ ; because the stiff $0°$ ply owns the negative slot and the soft $90°$ ply the positive one, and $Q_{11}^{(0°)} \neq = Q_{11}^{(90°)}$ , the two do not cancel. Pull in $x$ : the stiff bottom resists more than the soft top → the plate curves.
Single ply / zero thickness (degenerate). One layer with $h_{0} = - t /2, h_{1} = + t /2$ : then $[B] = \frac{1}{2} Q ((t /2)^{2} - (t /2)^{2}) = 0$ automatically (a single ply is its own mirror), $[A] = Qt$ , $[D] = Q t^{3} /12$ — recovering the familiar plate result. And if any $t_{k} \to 0$ , that ply simply drops out of all three sums. No division by anything, so nothing blows up.

Downstream this ABD feeds First Ply Failure, picks up temperature terms in Thermal and Hygroscopic Effects, extends to cores in Sandwich Panel Theory, and becomes an element in Finite Element Modeling of Composites.

The one-picture summary

The whole derivation in a single flow: a straight-line strain ( $ϵ^{0} + z κ$ ) enters each ply's $[Q]_{k}$ , producing a jumping stress; integrate $\int d z$ and $\int z d z$ to collect force ( $[A], [B]$ ), integrate $\int z d z$ and $\int z^{2} d z$ to collect moment ( $[B], [D]$ ), then stack the four blocks into the single $6 \times 6$ ABD matrix that maps loads ${N, M}$ to deformations ${ϵ^{0}, κ}$ . That one box is the laminate's complete stiffness passport.

Recall Feynman retelling — say it back in plain words

Imagine a sandwich of stiff sheets glued together, and draw a straight pin through it from bottom to top. When you tug and bend the sandwich, the pin stays straight but slides and tilts — sliding is stretch, tilting is bend, and the strain anywhere on the pin is "middle slide plus height times tilt." Each sheet pushes back with its own stiffness because its fibres point its own way, so the push-back (stress) has kinks where the sheets meet. To get the total sideways force, add up all that push-back over the height. To get the total twisting moment, add it up again but weight each slice by how far it is from the middle, because far-away pushes twist harder. Doing those two sums produces three tables of numbers: $[A]$ for stretching, $[D]$ for bending, and $[B]$ that couples the two — $[B]$ is zero when the sandwich is a perfect mirror about its middle, and nonzero when it is lopsided, which is exactly why a lopsided panel warps when you merely pull it. Glue all three tables into one $6 \times 6$ box and you have the laminate's complete instruction manual: feed it loads, it hands back deformations.

Recall

Where does zero of $z$ go, and why? ::: At the mid-plane, so symmetric stacks make $[B] = 0$ and $ϵ (0) = ϵ^{0}$ . What integral makes $[A]$ , $[B]$ , $[D]$ ? ::: $\int d z$ , $\int z d z$ , $\int z^{2} d z$ of $[Q]$ through the thickness respectively. Why does $[B]$ appear twice in the $6 \times 6$ ? ::: Both the force integral and the moment integral produce the same $\frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2})$ weight — one coupling matrix, two appearances. Physical meaning of $[B] \neq = 0$ ? ::: Pulling causes curving (and bending causes stretching); the stack is not mirror-symmetric about its mid-plane. What is $γ_{x y}$ physically? ::: Engineering shear strain — the total change of the right angle between $x$ and $y$ fibres, in radians.