A laminate is a stack of thin plies. We slice it with a vertical coordinate z, measured from the mid-plane (the surface at z=0), positive pointing "up". Look at Figure 1: each ply k occupies a slab between two heights.
Recall The reduced stiffness
[Q]k in one breath
[Q]k is the 3×3transformed reduced stiffness of ply k — the law σk=[Q]kϵ inside that ply. You get it by taking the ply's own-axis stiffness (built from E1,E2,G12,ν12) and rotating it by the fiber angle θk (see Reduced Stiffness Matrix Q and Transformation of Stiffness). Its form: [Q]k=Q11Q12Q16Q12Q22Q26Q16Q26Q66k. Two plies of the same material but different angle have different[Q]k.
Figure 2 shows how the three sub-matrices arise from integrating [Q]k against different powers of z — this is the geometric root of the scaling laws the questions probe.
Figure 3 shows why mirror symmetry forces [B]=0: the signed z-lever contributions of mirror-paired plies cancel.
[B]=0 is guaranteed by any laminate that has the same plies top and bottom.
False — you need mirror symmetry about the mid-plane (same material, angle, AND same distance ∣z∣, as in Fig. 3). A stack like [0/90/0/90] has matching plies but they are not mirror-placed, so [B]=0.
A symmetric laminate can still bend when you pull on it.
False — for a symmetric laminate, [B]=0 in Eq. (4), so a pure in-plane force N produces only mid-plane strain ϵ0 and zero curvature κ. Bend-stretch coupling is exactly the thing [B] carries, and it has vanished.
[A] depends on the order in which you stack the plies.
False — Eq. (1) weights each ply only by its thickness tk, not its position. Reshuffling the same plies leaves [A] identical; only [B] and [D] feel the order.
[D] depends on stacking order.
True — Eq. (3) weights each ply by 31(hk3−hk−13), which grows fast with distance from the mid-plane, so plies at the surface dominate. Moving a stiff ply outward raises [D].
A balanced laminate always has [B]=0.
False — "balanced" (equal numbers of +θ and −θ plies) kills A16 and A26, i.e. shear-extension coupling in [A]. It says nothing about mid-plane symmetry, so [B] can still be non-zero.
The [ABD] matrix is always symmetric.
True — because [B] appears in both the force row and the moment row of Eq. (4), the full 6×6 is symmetric (ABBD). This is a consequence of a single strain energy governing both, so mixed partials are equal.
If two plies have the same fiber material they have the same [Q]k.
False — [Q]k is the transformedreduced stiffness in laminate axes, so a ply's angle θk changes it. A 0° and a 90° ply of identical material have different [Q]k (see Transformation of Stiffness).
Adding a thin ply far from the mid-plane changes [D] more than adding it near the mid-plane.
True — the z3 weighting in Eq. (3) means a ply at large ∣z∣ contributes hugely to bending stiffness, while one at z≈0 barely moves [D] (its hk3−hk−13 is tiny).
The claim confuses which matrix vanishes. Symmetry gives [B]=0, not[A]=0; by Eq. (1) [A] is a sum of positive-definite [Q]ktk terms and is never zero for real plies.
"[B] has units of stiffness, same as [A]."
Wrong units. [A] is Pa·m, [B] is Pa·m2 (it multiplies curvature κ in 1/m to give force/width N). You cannot add or compare [A] and [B] entries directly.
"To get strains, just multiply the loads by the [ABD] matrix."
Eq. (4) reads [NM]=[ABD][ϵ0κ], so to go from loads to strains you must invert: multiply by [ABD]−1.
"For a symmetric laminate under Nx only, I still need the full 6×6 inverse."
When [B]=0 Eq. (4) is block-diagonal, so ϵ0=[A]−1N and κ=[D]−1M decouple. With M=0 you only invert the 3×3[A].
"∫hk−1hkzdz vanishes for every ply, so [B] is always zero."
That integral only vanishes for a ply straddling the mid-plane symmetrically; for a ply lying entirely on one side, 21(hk2−hk−12)=0. [B] in Eq. (2) is a sum over plies, and it vanishes only when those signed contributions cancel.
"Curvature κ and strain ϵ0 have the same units, so [A] and [D] are interchangeable."
No — ϵ0 is dimensionless while κ has units 1/m. That difference is exactly why [A] (Pa·m) and [D] (Pa·m3) differ by two powers of length.
"The [0/90] two-ply laminate is symmetric because it has one 0 and one 90."
It is antisymmetric, not symmetric — the mirror of the 0° bottom ply would be a 0° top ply, but the top is 90°. Hence [B]=0 and pulling causes curvature.
Why does the same integral 21(hk2−hk−12) show up in both the force and moment equations?
In N it comes from integrating zκ (curvature's effect on force); in M from integrating zϵ0 (stretch's lever-arm moment). The shared integral of Eq. (2) is precisely why one matrix [B] couples both directions.
Why is the Kirchhoff-Love assumption (ϵ(z)=ϵ0+zκ) linear in z?
"Plane sections stay plane and perpendicular" means a straight fiber through the thickness stays straight after loading, so its strain varies linearly with height z — no bulging, no shear warping.
Why does [D] behave like the beam quantity EI?
Both measure resistance to bending, and both get their strength from material stiffness times the square-ish moment of area — here the z3 integration in Eq. (3) is the tensor cousin of ∫z2dA that makes up I.
Why can we integrate ply-by-ply instead of doing one continuous integral?
Each ply has a constant[Q]k through its own thickness, so the through-thickness integral splits into a sum of clean per-ply pieces (Fig. 2). Discontinuous stiffness forces the summation form.
Why does an asymmetric laminate warp when it merely cools down, even with no applied load?
Uneven ply placement gives [B]=0, so the thermal contraction strains (a stretching effect) feed into curvature through [B] — the same coupling covered in Thermal and Hygroscopic Effects.
Why do engineers prefer symmetric layups for spacecraft panels?
[B]=0 decouples stretch and bend, so thermal cycling in orbit produces no unwanted warping and the panel stays flat — critical for mirror mounts and antennas.
A single isotropic ply centered on the mid-plane: what is [B]?
Zero — one ply symmetric about z=0 has 21(h12−h02) with h1=−h0, so the term in Eq. (2) vanishes. A lone centered ply is trivially symmetric.
What happens to [A], [B], [D] if you double every ply thickness uniformly (scale total thickness by 2)?
[A] scales by 2 (linear in t, Eq. 1), [B] by 4 (h2, Eq. 2), and [D] by 8 (h3, Eq. 3). Thicker laminates gain bending stiffness fastest.
A laminate with all plies at the same angle (e.g. [0/0/0]): is it a "laminate" in the coupling sense?
Effectively no coupling from angle differences — every [Q]k is identical, so it behaves like a single thick orthotropic plate and [B]=0 (it is symmetric about its own center).
What is [B] for a laminate that is symmetric in geometry but the two halves are different materials placed as mirror images?
Still [B]=0 — the mirror-image condition requires matching [Q]k AND matching ∣z∣; if both halves mirror exactly (same material at mirrored positions), symmetry holds and [B] vanishes.
Zero applied moment but a non-symmetric laminate under pure Nx: is the curvature zero?
No — with [B]=0, M=0 does not force κ=0; solving the coupled Eq. (4) generally gives non-zero κx. The load bends it even though you applied no moment.
When is the [ABD] matrix ill-conditioned (hard to invert accurately)?
When entries span wildly different magnitudes — [A]∼Pa·m and [D]∼Pa·m3 differ by the thickness squared, so mixing them in one 6×6 can produce large condition numbers, especially for thin laminates.
Does a laminate that fails First Ply Failure suddenly change its [ABD] matrix?
The linear ABD is computed from intact ply stiffnesses; predicting First Ply Failure uses ABD to get ply stresses, but ABD itself only updates if you degrade the failed ply's [Q]k in a progressive-damage model.
Can classical laminate theory (and thus [ABD]) handle a thick sandwich with a soft core?
Poorly — CLT assumes thin plies with no transverse shear; a soft-core sandwich shears through the core, so you need Sandwich Panel Theory or a shear-deformable / FE approach.